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Lecture Comments (4)

1 answer

Last reply by: Taylor Wright
Wed Jun 12, 2013 10:41 PM

Post by Jose Gonzalez-Gigato on February 3, 2012

As always, great lesson, Ms. Pyo. I do have one note: in Example III, I believe that the computation of the midpoint of segment RS is incorrect. For the y coordinate it should be: (6+(-2))/2 resulting them in (0,2) and not (0,4).

1 answer

Last reply by: Mary Pyo
Fri Feb 3, 2012 11:31 PM

Post by Dro Mahmoudi on November 13, 2011

you are the best, you and educator really saved me from a big nightmare

Special Segments in Triangles

  • Perpendicular Bisector: A line or line segment that passes through the midpoint and is perpendicular to that side
  • Perpendicular Bisector Theorems:
    • Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment
    • Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment
  • Median: A segment that connects a vertex of a triangle to the midpoint of the side opposite the vertex
  • Altitude: A segment with one endpoint at the vertex and the other on the side opposite that vertex so that the segment is perpendicular to the side
  • Angle Bisector: A segment with one endpoint on the vertex and the other on the side opposite so that it divides the angle into two congruent angles
  • Angle Bisector Theorems:
    • Any point on the bisector of an angle is equidistant from the sides of the angle
    • Any point on or in the interior of an angle and equidistant from the sides of an angle lies on the bisector of the angle

Special Segments in Triangles

Find the point D on AB , so that CD is the median to AB .
  • A(3, 4), B( − 3, 2)
  • the midpoint of AB is point D ([(3 − 3)/2], [(4 + 2)/2])
D(0, 3).
, any point on AD is equidistant from AB and AC , determin what kind of segment is AD .
  
Angle bisector.
Draw and label to a figure to illustrate AM is the altitude of ∆ABC , M is on BC .
  
Draw the median AD of ∆ABC on the coordinate plane.
  
Determine whether the following statement is true or false.
Any point on the median of a segment is equidistant from the endpoint of the segment.
  
False.
, DE is the perpendicular bisector of BC , EC = 3x + 2, EB = 5x + 6, find x.
  • EC = EB
  • 3x + 2 = 5x + 6
  • − 2x = 4
x = − 2

∆ABC, A(3, 4), B(0, − 1), C(4, − 3), AD is the altitude of BC , find the line passes through points A and D.
  • the slope of BC is : [( − 3 − ( − 1))/(4 − 0)] = − [1/2]
  • the slope of AD is 2
  • then line AD is: y = 2x + b
  • line AD passes through A(3, 4)
  • 4 = 2*3 + b
  • b = − 2
line AD: y = 2x − 2.
Fill in the blank of the statement with always, sometimes or never.
Any point equidistant from the endpoints of a segment _____ lies on the perpendicular bisector of the segment.
Always

Given: AB = AC , AD is the median to BC
Prove: AD is also the altitude of BC .
  • Statements; Reasons
  • AB = AC; Given
  • AD is the median of BC; given
  • BD ≅ CD ; definition of median
  • AD ≅ AD ; refelxive prop ( = )
  • ∆ABD ≅ ∆ACD ; SSS
  • ∆ ADB ≅ ∆ ADC ; CPCTC
  • m∠ADB = m∠ADC ; definition of ≅ ∠s
  • m∠ADB + m∠ADC = 180o ; definition of linear angles
  • 2 m∠ADB = 180o ; substitution prop of ( = )
  • m∠ADB = 90o ; division prop of ( = )
  • AD ⊥BC ; definition of perpendicular
  • AD is also the altitude of BC ; definition of altitude.
Statements; Reasons
AB = AC; Given
AD is the median of BC; Given
BD ≅ CD ; definition of median
AD ≅ AD ; refelxive prop ( = )
∆ABD ≅ ∆ACD ; SSS
∆ ADB ≅ ∆ ADC ; CPCTC
m∠ADB = m∠ADC ; definition of ≅ ∠s
m∠ADB + m∠ADC = 180o ; definition of linear angles
2 m∠ADB = 180o ; substitution prop of ( = )
m∠ADB = 90o ; division prop of ( = )
AD ⊥BC ; definition of perpendicular
AD is also the altitude of BC ; definition of altitude.
Determine the following statement is true or false.
The median to the base of an isosceles triangle also bisects the vertex angle and is perpendicular to the base of the triangle.
True

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Special Segments in Triangles

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Perpendicular Bisector 0:06
    • Perpendicular Bisector
  • Perpendicular Bisector 4:07
    • Perpendicular Bisector Theorems
  • Median 6:30
    • Definition of Median
  • Median 9:41
    • Example: Median
  • Altitude 12:22
    • Definition of Altitude
  • Angle Bisector 14:33
    • Definition of Angle Bisector
  • Angle Bisector 16:41
    • Angle Bisector Theorems
  • Special Segments Overview 18:57
    • Perpendicular Bisector
    • Median
    • Altitude
    • Angle Bisector
    • Examples: Special Segments
  • Extra Example 1: Draw and Label 22:36
  • Extra Example 2: Draw the Altitudes for Each Triangle 24:37
  • Extra Example 3: Perpendicular Bisector 27:57
  • Extra Example 4: Draw, Label, and Write Proof 34:33

Transcription: Special Segments in Triangles

Welcome back to Educator.com.0000

This next lesson is on special segments within triangles.0002

The first one (there are a few) is the perpendicular bisector.0007

Now, we know that "perpendicular" means that it is going to form a right angle.0014

And a "bisector" is a segment or an angle that cuts a segment or an angle in half.0021

So, a perpendicular bisector is going to be a line or a line segment that passes through the midpoint0030

(that is the bisector part of it) and is perpendicular to that side.0039

Two things: the segment has to pass through the side so that it is going to be perpendicular, and it is going to bisect that side.0045

So, if I want to draw the perpendicular bisector of the side AC, I have to draw a line or a line segment0055

that is going to be perpendicular to the side and is going to bisect it--it is going to cut it in half at its midpoint.0065

Let's say that that is the midpoint right there; that means that this is cut in half.0073

And then, I have to draw a segment that is going to be perpendicular and is going to bisect it--something like that,0079

where it is perpendicular and it bisects; so this right here is going to be the perpendicular bisector.0090

So again, the perpendicular bisector is a line or a line segment that is going to make it perpendicular and bisect the side.0098

Now, I could draw a perpendicular bisector for each side of the triangle; this is just for side AC.0106

But since I have three sides, I can draw three perpendicular bisectors: one for each side.0114

If you are only required to draw one, then you can just draw it like this.0121

You don't have to draw it all the way through; it can be a line or a line segment, so you can draw a line,0126

or you can just cut it right here and just make it a segment.0131

But that would just be for one of the sides; you can draw three, for each of the sides.0136

If you wanted to draw a perpendicular bisector for side AB, it might be helpful for you to just turn your paper so that this is the horizontal side.0142

So again, we want to have the midpoint, and then I am going to do that to show that that is the midpoint.0154

And then, I need to draw a segment that is going to be perpendicular, like that.0164

That is two; the third one is going to be for side BC.0179

So again, ignore this line right here; so then, that would be 1, 2, 3, to show that those two parts are congruent.0183

And then, draw...like that.0200

What should happen is that all three perpendicular bisectors, the ones that you draw for each of the sides, should all meet at one point.0217

Again, we have a perpendicular bisector; it is a line or line segment that is perpendicular to the side and bisects the side.0229

That means that it cuts it in half; that is a perpendicular bisector.0242

And then, for the perpendicular bisector, a couple of theorems: Any point on the perpendicular bisector of a segment...0249

when you just draw another triangle, and let's say my perpendicular bisector is right there,0257

then what this theorem is saying is that any point on the perpendicular bisector0281

(now, this is the perpendicular bisector)--any point on this line is equidistant from the endpoints of the segment.0284

So, this is the segment, AB; I could pick any point on this perpendicular bisector, and the distance from this point to this endpoint,0294

and this point to this endpoint, is going to be the same.0307

"Equidistant" means that it is the same distance; from this point to point A and this point to point B is going to be the same; that is what it is saying.0310

Or I could pick any point, maybe here; again, from this point to point A, and this point to point B, this is going to be the same.0322

That is what this one is saying.0334

And then, the next one: Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment.0335

This is just the converse of this; it is saying that if, without knowing where my perpendicular bisector is,0342

I draw two segments from point A and from point B out to the same point,0349

so that the two distances are the same, then it will lie on the perpendicular bisector.0356

The first one is saying that, if the point lies on the perpendicular bisector, then it is equidistant from the two endpoints, point A and point B.0362

The second one is saying that, if I draw two segments to a point equidistant from point A and point B, then it lies on the perpendicular bisector.0373

OK, so then, that was the first one; that was the perpendicular bisector; the second one is a median.0392

Now, when you think of median, think of middle or midpoint.0398

It is kind of like a perpendicular bisector; with perpendicular bisectors, we worked with the midpoint, too.0406

But that had two conditions; it had to be perpendicular to the side, and the bisector had to go through the midpoint of the side.0413

Median is just through the midpoint--just that one condition.0422

So, think of median as "middle"; but the condition here is this--it is going to go to the middle of the side, but from the vertex opposite that side.0426

So, the vertex opposite this side is B; it can't be A, and it can't be C; it has to be B.0449

So, when I draw a segment from the vertex to that point, this would be the median.0458

If I label this as D, BD is the median of this triangle, of this side, AC.0473

Again, the difference between this and a perpendicular bisector: a perpendicular bisector has nothing about the vertex.0481

You don't care about where the vertex is.0489

You just have to draw the segment so that it is perpendicular to the side, and it is cutting at the midpoint.0492

The median is just the segment from the vertex to the midpoint of the side opposite.0498

So then, again, since we have three sides of a triangle, I can draw three medians in a triangle.0509

This is just one median; the next median I can draw from...let's say that is the midpoint--I'll do that.0514

So again, I am going to draw it from this point, the segment with two endpoints; one is there, and one is at this vertex.0523

And then again, from this one, let's say it is about here: 1, 2, 3, 1, 2, 3...to show that these parts are the same, or congruent.0540

And then again, I am going to draw from there all the way to the vertex opposite.0554

And again, for a median, if you draw all three medians of a triangle, then it should meet at one point, right there.0566

That is the median; the median is like the middle; so far, we did "perpendicular bisector," and we did "median."0575

OK, let's do this problem for median: Find the point S on segment AB so that CS is a median.0583

I want to find a point labeled S on AB, this side right here, so that from C to that S is going to be a median.0597

Now, remember: median has to do with the midpoint or middle.0608

Remember: the median has two endpoints: one is from the vertex, so that means it is going to come from C, vertex C;0615

and it is going to go to the midpoint of AB, and that point is going to be labeled S.0623

So, I need to find the midpoint of AB, because CS has to be a median; that means S has to be the midpoint of AB.0630

How do I find the midpoint when I am given two points?0641

A is at 1, 2, 3...(-3,2); B is (1,-4); so I am just going 1, 2, 3, 4; positive 1, negative 4.0644

To find the midpoint of this point to this point, it is going to be (x1 + x2) divided by 2;0666

so I am going to add up the x's and divide it by 2; and then I am going to add up my y-coordinates and divide it by 2.0676

So, it is like the average; to find the midpoint, you are going to find the average of the x's and the average of the y's.0684

So then, for the x's, it is (-3 + 1)/2; for the y's, it is (2 + -4)/2; so this would be -2/2 and -2/2.0690

Well, -2/2 is -1, and -2/2 here is -1; that means (-1,-1).0709

This point right here is where S is; that means that, if I draw a line from C all the way to S, like that, that is a median,0717

because S is the midpoint between A and B; so here is point S.0731

The third segment is the altitude: now, altitude--just think of it as being perpendicular to the side.0744

It is kind of like the perpendicular bisector, except that there is no bisector--just the "perpendicular."0763

But then, one of the endpoints also has to be at the vertex.0769

One point is at the vertex, and one point on the side opposite, so that it is perpendicular.0777

A median only takes the "midpoint" side, and the altitude only takes the "perpendicular" side.0783

And then, both of them together is like the perpendicular bisector.0788

So, I just have to draw a segment going from C down to this side so that it is perpendicular, not caring about midpoint.0792

All I care about is that it is from this endpoint at the vertex, and it is perpendicular to the side.0804

Let's say...that right there; now again, there are three sides, so I need to draw three altitudes.0814

Now, it kind of looks like this BC is already the altitude; I'll just draw it like that.0829

And again, it is from this vertex to this side, so that it is perpendicular.0840

And then, where are they meeting?--right there.0855

So again, an altitude is from the vertex to the opposite side so that the segment is perpendicular.0858

This is the fourth one, the angle bisector; now, the angle bisector is bisecting (cutting in half),0875

but it is the angle that is being cut in half, not the segment, like the perpendicular bisector.0886

An angle bisector is a segment with one endpoint on the vertex, and it is coming out;0891

but for this one, because it is the angle bisector, we don't care where it lands on the side.0901

It doesn't matter where it lands; it is not going to be perpendicular; it is not going to be the midpoint...0907

it could be, but that is not what it has to be.0912

All it has to be: the condition is that it is coming from the vertex, and it is cutting the angle in half.0916

"Angle bisector" means that the angle is being cut in half.0924

So, all I care about is making sure that I draw a segment so that this angle is going to be bisected.0928

OK, well, let's just say that that is cut in half.0939

And then, if I draw something like that, let's say that is cutting in half the angles.0943

And then, if I draw that, we can say that this angle is cut in half.0954

See how this has no regard for where it is touching the side (as long as it is touching it).0969

But I am not saying that this has to be perpendicular, or it has to be at the midpoint--none of that.0976

The only thing is that the angle has to be bisected.0983

And again, I do three because there are three sides; I have to do it to each of those angles.0986

And then again, they meet at one point.0992

That is the angle bisector, where the segment is bisecting the angle from the vertex.0995

Any point on the bisector of an angle (let me draw a triangle again; I'll say my angle bisector is right there) is equidistant from the sides of the angle.1004

If I just have a point that is any point on this angle bisector, it is equidistant to the sides, like that, or like that.1031

Remember: if we want to find a distance from a point to a side, it has to be perpendicular.1042

Remember: if you are standing in front of a wall, how do you find the distance--how far away you are from that wall?1048

You don't measure at an angle; you have to measure directly, so that you are perpendicular.1053

That distance from you to the wall has to be perpendicular to the wall.1058

You can't just turn your body at an angle and find your distance that way.1064

The same thing works here: if you want to find the distance between this point and this side, this is you; this is the wall.1070

It has to be perpendicular; so let me draw this out again so that it looks like it is perpendicular.1075

You are going to go straight out like this; it is just saying that, if this is the angle bisector,1086

then any point on the angle bisector is equidistant; that means that the distance to the sides of the angle,1092

which are these two sides, is going to be equidistant.1102

And then, this is the converse; any point on, or in the interior of, an angle, and equidistant from the sides of the angle, lies on the bisector of the angle.1106

It is the same thing; it is just saying that, if I just find the distance to a point from the sides,1116

so that it is equidistant, then it is going to lie on the angle bisector.1125

On the angle bisector, any point is going to be equidistant from the sides.1131

OK, let's go over the four that we went over, the special segments of a triangle.1139

A perpendicular bisector: remember: it had to be perpendicular, and bisecting the side; that is the perpendicular bisector.1147

For a median, if this is the side of a triangle, then we don't care what this looks like, as long as this is bisected--the midpoint of the side.1173

The altitude: we don't care what these sides look like, as long as it is perpendicular.1190

The angle bisector: it is coming out from the angle of the triangle so that these are bisected.1204

A perpendicular bisector is going to be like this, like this, and like this; you can draw arrows or not; that is a perpendicular bisector.1221

For a median, it is from the vertex, so that it is congruent; from the vertex...that is the median.1256

The altitude is just drawn so that it is perpendicular, so it is like that.1286

And the angle bisector, the last one, is drawn so that it is bisecting the angles.1311

Now, this is supposed to be bisecting the angle.1331

Those are the four special segments.1338

Again, the perpendicular bisector has to be perpendicular and bisect the sides.1341

The median is just bisecting the side; the altitude is just perpendicular; the angle bisector is bisecting the angle.1346

Let's do our examples: Draw and label a figure to illustrate each: BD is a median of triangle ABC, and D is between A and C.1356

BD is the median, so I am going to draw triangle ABC; there is A, B, C; and BD--that means that it is coming out from here.1370

BD is the median of triangle ABC, and D is between A and C.1385

That means that, since we are dealing with BD as the median, D has to be the midpoint; there is D; there is BD.1391

And all you had to do is draw it and label it.1405

The next one: GH is an angle bisector of triangle EFG, and H is between E and F.1411

Triangle EFG: H is between E and F; GH is an angle bisector; so H is between E and F so that GH is an angle bisector.1421

Now, we don't care, as long as H is anywhere in between here; it is not going to be perpendicular;1444

it could be, but that is not the rule; the rule is that the angle is bisected.1453

This is like this, and then this point would be H.1461

So, GH is an angle bisector of that triangle.1471

Draw the altitudes for each triangle: I want to draw an altitude (remember: the altitude also has an endpoint on the vertex).1478

So, if I want to draw an altitude from A to side BC, it looks like this is already the altitude.1492

I can just say that this side right here would be the altitude of this side BC.1499

And then, B to AC is going to be like that, and then C to AB is going to be like...that is not right...let me erase part of my triangle...1503

so then, this BC would be the perpendicular bisector of AB, also.1527

This is going to be the altitude of BC; and then, CB is going to be the altitude of AB.1536

And then, for this one, this triangle is an obtuse triangle because angle B is greater than 90 degrees.1551

So, if I want to draw the altitude from this to the side, that is pretty easy; that just goes straight down.1559

But then, this one is a little bit different, because I obviously can't draw an altitude1570

from point A to somewhere between B and C, so that it will be 90 degrees.1577

So, what I have to do is extend this out a little bit (it is kind of hard to draw a straight line on this thing).1584

Let's say that that is my straight line; that is CB extending out.1595

Then, my altitude will have to go outside of this triangle, because, since it is an obtuse angle,1603

I would have to draw it on the outside so that it will be 90 degrees.1616

If I draw it on the inside, it is just going to be a bigger obtuse angle.1619

That is the altitude for that side; and then, if I want to draw the altitude from C to this side, AB, the same thing: it is an obtuse angle.1628

So, I am going to extend this out; I am going to draw from C all the way there so that it is perpendicular.1639

Those would be my three altitudes; now, if you want them to all meet, which they should, in this case, they all met right here.1651

For this one, you would just have to keep drawing this out, keep drawing this out,1659

and keep drawing this out, and then they would eventually meet right there.1664

But if you just have to draw the altitudes, then you would have to just draw that, draw that, and then draw this.1668

The coordinate points of triangle RST are those three points; AB is a perpendicular bisector, so it is this, through RS.1679

So now, I don't have to actually graph this out on a coordinate plane.1698

But if you are a visual person, and you like to see how it looks, then you can go ahead and plot them.1706

I am just going to do a little sketch of what it will look like.1715

So, if I have -2 right there, let's say this is R.1721

S, let's say, is -2; (2,-2) is S; and T is (5,4), so here is T.1727

So, my triangle is going to look something like this.1741

AB is a perpendicular bisector through RS; that means that my perpendicular bisector is through this side.1749

That means that this side is going to be the one that is perpendicular to it and that is bisected.1758

So, that is the perpendicular bisector, which means that it is going to look something like that; it is perpendicular.1765

And then, this is going to be A and B, this point and this point.1784

Find the point of intersection of AB and RS.1796

I want to find the point of intersection; now, first of all, we have to find the point of intersection between AB and RS.1805

And then, I want to find the slope of AB and RS.1817

So, the point of intersection, we know, is right there; we also know that the same point, that point A, is the midpoint of RS.1822

So, as long as I find the midpoint of RS, that would be the point of intersection between the two segments,1833

because again: this segment and this segment meet at point A, which is the midpoint of RS.1839

The point of intersection is going to be the midpoint.1846

#1: the midpoint of RS--I am going to use these two points; remember, to find the midpoint, I am going to add up my x's,1852

divide it by 2, and add up my y's...6 - -2, divided by 2; it is going to be 0, comma...minus a negative becomes a plus,1864

so it is plus 2, is 8; 8 divided by 2 is 4; so then, #1: the point of intersection would be (0,4).1881

And again, the reason why I did midpoint is because that is where they intersect: they intersect at the midpoint of RS.1892

And then, that is my answer; that is the point of intersection.1900

Then, #2: Find the slope of AB and RS.1903

I have points R and S; so to find the slope, it is (y2 - y1)/(x2 - x1); this is slope.1908

Again, using the same points, let's see: if I label this (x1,y1),1924

and label this as (x2,y2)...again, the x2 and y21933

is not talking about "squared"; make sure you write this 1 and these 2's below, not above it like an exponent.1942

And this is just saying the first and the first y; the second x and second y, because we know that this point is (x,y), and this is also (x,y).1958

So, these are the first (x,y)'s, and these are the second (x,y)'s.1967

So, y2 - y1 is -2 - 6 (and this is the slope of RS);1970

and then, x2 is 2, minus -2, which is -8 over...this becomes plus, so it is 4; this is -2.1984

The slope of RS is -2; now, I have to also find the slope of AB.1998

I don't know the point for B, but I don't have to know, because, if you have two lines,2009

and they are perpendicular to each other, then, remember: their slopes are negative reciprocals of each other.2020

So, now that I know the slope of RS, to find the slope of AB, it is just the negative reciprocal of it.2026

So, this is the slope of RS, and then the slope of AB is going to be the negative reciprocal; so that is the negative of (-1/2).2033

So then, this is going to be positive 1/2.2053

One is -2, and the other one is positive 1/2.2065

OK, the fourth example: Draw and label the figure for the statement; then write a proof.2074

The median to the base of an isosceles triangle bisects the vertex angle.2079

I am going to draw and label a figure, so I need an isosceles triangle.2084

Let's say that this is my isosceles triangle; the median to the base of it...we know that these are congruent, because it is the median...2097

of an isosceles triangle...bisects the vertex angle; that means that we want to prove...2116

Let me just label this; now, since this is how you want to draw and label it, you can just draw and label it however you want.2123

So, if I label this ABC, I can label this point as D.2130

So, my given statement--what do I know?--what is given?2138

I have an isosceles triangle, so I can say that triangle ABC is isosceles.2147

And then, I can say that BD is a median, because those are parts of the information that I have to use.2162

Now, I want to prove that it bisects the vertex angle; I am going to prove that angle ABD is congruent to angle...2176

if I said ABD, then I have to say angle CBD, because A and C are corresponding; so if I say ABD, then I have to say CBD.2197

OK, from here, I am going to do my proof: so my statements and my reasons...#1: Triangle ABC is isosceles,2211

and then BD is the median; and the reason for that is because it is given.2240

2: From here, I can say that AB, this side, is congruent to this side.2250

Now, let's see what we have to do: I am trying to prove that this angle right here is congruent to this angle right here.2259

Now, in order for me to prove that those two angles are congruent,2273

I would probably have to first prove that these two triangles are congruent,2280

because there is no way that I could just say that this angle is congruent to this angle.2285

But if I prove that these two triangles are congruent, then I can say that any two corresponding parts are congruent.2289

So then, once these two triangles are congruent, then these two angles can be congruent.2297

As long as they are corresponding, any two parts of the triangles are congruent.2303

Then, I have to focus on how I am going to prove that these two triangles are congruent.2308

Well, I know that these two sides are congruent; I know that these two sides are congruent.2313

And I can say that this side of this triangle is congruent to this side of this triangle; that is the reflexive property.2319

I can prove that these two triangles are congruent by SSS; if you remember the rules, there is SSS, SAS, AAS, and then Angle-Side-Angle, ASA.2330

So, I could do that, or I have another option; I can say that, because (remember) in an isosceles triangle,2344

if I have the two legs being congruent, then these two angles are also congruent--remember: the base angles are also congruent.2352

So, I can say that, too, and then prove that these two triangles are congruent through SAS.2360

Either one works; the important thing is that we prove that these two triangles are congruent,2366

so that we can say that these two angles, those two parts, are congruent.2372

It is up to you--do it however you want to do it.2379

I am just going to use the reflexive property, and say that this side is congruent, and use SSS.2382

So, I am going to say that AB is congruent to CB, and that is my side; the reason would just be "definition of isosceles triangle,"2390

because the definition of isosceles triangle just says that two legs are congruent--"two or more sides of a triangle are congruent."2405

And then, I am going to say that these two parts are congruent.2419

So, even though I have it shown on my diagram, I have to write it as a step.2422

AD is congruent to CD, and the reason for that--I am going to say "definition of median,"2430

because it is the median that made those two parts congruent--so it is just "definition of median."2443

And then, that is another side that I have; and then I can say, "BD is congruent to BD," and this would be BD of this triangle,2451

and this would be BD of the other triangle; so I am saying that a side of one triangle is congruent to a side of another triangle.2463

And that would be the reflexive property.2469

Now, if you chose to say that angle A is congruent to angle C, then you can say "isosceles triangle theorem" as your reason--2476

"isosceles triangle theorem" or "base angles theorem," because,2484

since that is an isosceles triangle, automatically the base angles are congruent.2488

And then, the fifth step...that would be another side, so then, your reason,2493

if you say that the triangles are congruent here (the next step), wouldn't be SSS, like mine would be; it would be SAS.2499

Triangle ABD is congruent to triangle CBD, and again, my reason is SSS.2510

Then, from there, I can say that angle ABD is congruent to angle CBD; what is my reason?2525

Well, see how here you proved that the triangles are congruent.2536

Then, once this is stated, then you can say that any two parts are congruent by "corresponding parts of congruent triangles are congruent," CPCTC.2541

So, that would be my sixth step.2558

Again, in order to prove that these two angles are congruent, because there is no direct way to do it,2561

I have to prove that these two triangles are congruent so that I can say2568

that two corresponding parts, those two angles, are going to be congruent.2573

And then, you can do that by Side-Side-Side or Side-Angle-Side.2577

And then, once you prove that the triangles are congruent, then you can say that those two angles are congruent.2581

When you draw and label, if it doesn't give you a figure or a diagram for it, then just draw your own.2587

You can label it how you want, and then that would base your given and your prove statement.2594

But as long as you write a proof for this statement that they give you, "The median to the base of an isosceles triangle2600

bisects the vertex angle," this would be your conclusion (your "prove" statement).2608

Your hypothesis is going to be your given, and your conclusion of your statement is your "prove" statement.2612

Well, that is it for this lesson; I will see you next time.2622