- anonymous

(Geometry)
Can someone justify this for me?

- jamiebookeater

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- anonymous

So this is the problem statement:
"Suppose ΔABC is an acute triangle. Let the bases of the altitudes to B and C be E and F, respectively, and let A' be the midpoint of BC. Prove ΔA'EF is isosceles."
|dw:1437180207621:dw|
Okay, so my professor when hinting at the solution to this problem said to draw a circle for which the length BC is the diameter and that doing so should show that the right angles are inscribed on the circle. But I don't know how we can justify that is true. Sure, you can draw it and see it, but what proof is there to say that A'F and A'E would be radii and that those right angles would be inscribed?

- Mertsj

Do you mean that E & F are the bases of the altitudes FROM A & B?

- sdfgsdfgs

Converse of Thales theorem:
https://en.wikipedia.org/wiki/Thales%27_theorem
It shows that a circle containing triangle BEC will have its diameter on BC. Since A' is the mid-point, it is the center of the circle.
Apply the same reasoning for triangle BFC, it can be shown that the two circles are the same one. Hence A'F and A'E must be equal in length.

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- anonymous

Oh, I apologize, you may be right, Mertsj. I think I drew the picture wrong x_x

- anonymous

What would the appropriate picture be then? I drew the picture on another question that was written almost exactly the same, but I didnt realize there was a single word change, so yeah, changes everything.

- Mertsj

If BC is a diameter, then A' is the center of the circle and E and F cannot be points on the circle.

- Mertsj

Well, it would help if you would post the problem EXACTLY as it is written so we could get a proper diagram.

- anonymous

No, I posted the probably exactly as written, I just drew the picture wrong.

- anonymous

*the problem

- sdfgsdfgs

@Mertsj why E n F cannot be points on the circle?

- Mertsj

I don't think so because B and C are vertices of the triangle. One cannot draw altitudes TO a vertex. The definition of the altitude of a triangle is a line that extends from one vertex of a triangle perpendicular to the opposite side. You start at the vertex.

- anonymous

Lemme see if I can get a picture then to show.

- sdfgsdfgs

But u start at B which is a vertex, draw a perpendicular line to AC. u get pt E. BEC forms a right-angled triangle, then u draw a circle inscribing it....

- triciaal

why wouldn't "an altitude to a vertex" be a perpendicular bisector ?

- anonymous

http://i201.photobucket.com/albums/aa120/ApocalypticDeity/geo_zpsy7tgay7j.png

- triciaal

using the information of a circle then the angle edge is 1/2 the angle at the center here BC is the chord through the center therefore the diameter angle 180 at the edge = 90

- Mertsj

|dw:1437183637273:dw||dw:1437183930649:dw|

- triciaal

using similar figures
CBE = CFB
AFE similar to ABC

- sdfgsdfgs

@Mertsj yup yup - so w the 2 right-angled triangles, BEC and BFC and the same inscribing circle with center at A', A'F=A'E right?

- anonymous

RIght, thats what I thought the picture would be. But I still wasnt sure why those right angles would be inscribed. I have to write it out as a proof, so I didnt want to just draw the circle and make the claim without knowing the reasoning.
@triciaal How do you show AFE is similar to ABC?

- sdfgsdfgs

@Concentrationalizing Converse of Thales theorem states that the inscribing circle of a right-angled triangle must have the hypothesis as its diameter:
https://en.wikipedia.org/wiki/Thales%27_theorem

- anonymous

I have to head out, but Ill be back to finish the discussion asap, hopefully real soon. Thanks guys :)

- triciaal

A'B = A'C so FE is parallel to BC

- triciaal

" Let the bases of the altitudes to B and C be E and F, respectively"
to me this is the perpendicular bisector
BF = FA

- Mertsj

|dw:1437185461552:dw|

- anonymous

So, I'm guessing that the circle in which that is true is unique? I'm sure it is, but just want to be sure.
And if the wording was incorrect, how should have it been phrased? Does the wording imply perpendicular bisectors like @triciaal suggests?

- Mertsj

|dw:1437186037868:dw|

- anonymous

Yeah, saying the points must be on the circle works then. I was thinking that there might be multiple ways to draw such a circle, but it seems like the circle that you would draw is unique and must fit those angles. Alrighty, that works then, thanks peoplez :)

- Mertsj

yw

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