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### 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

- A(3, 4), B( − 3, 2)
- the midpoint of AB is point D ([(3 − 3)/2], [(4 + 2)/2])

Any point on the median of a segment is equidistant from the endpoint of the segment.

- ―EC = ―EB
- 3x + 2 = 5x + 6
- − 2x = 4

∆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

Any point equidistant from the endpoints of a segment _____ lies on the perpendicular bisector of the segment.

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 = 180
^{o}; definition of linear angles - 2 m∠ADB = 180
^{o}; substitution prop of ( = ) - m∠ADB = 90
^{o}; division prop of ( = ) - ―AD ⊥―BC ; definition of perpendicular
- ―AD is also the altitude of ―BC ; definition of altitude.

―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 = 180

^{o}; definition of linear angles

2 m∠ADB = 180

^{o}; substitution prop of ( = )

m∠ADB = 90

^{o}; division prop of ( = )

―AD ⊥―BC ; definition of perpendicular

―AD is also the altitude of ―BC ; definition of altitude.

The median to the base of an isosceles triangle also bisects the vertex angle and is perpendicular to the base of the triangle.

*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

### Geometry Online Course

### 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 midpoint*0030

*(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 segment*0055

*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 bisector*0281

*(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 (x _{1} + x_{2}) 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 altitude*1570

*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 (y _{2} - y_{1})/(x_{2} - x_{1}); this is slope.*1908

*Again, using the same points, let's see: if I label this (x _{1},y_{1}),*1924

*and label this as (x _{2},y_{2})...again, the x_{2} and y_{2}*1933

*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, y _{2} - y_{1} is -2 - 6 (and this is the slope of RS);*1970

*and then, x _{2} 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 say*2568

*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 triangle*2600

*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

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