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 January 31st, 2010, 07:44 PM #1 Member   Joined: Jan 2010 Posts: 43 Thanks: 0 Geometry Triangle Proof If the altitude from vertex A in triangle ABC is also the bisector of angle A, prove AB=AC. I'm trying to do this without drawing a picture but I struggle with proofs. Any help or hints to get me started are greatly appreciated. I am using Birkhoff's Axiomatics. Thanks!
January 31st, 2010, 08:51 PM   #2
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From: Lexington, MA

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Re: Geometry Triangle Proof

Hello, meph1st0pheles!

Quote:
 $\text{If the altitude from vertex }A\text{ in }\Delta ABC\text{ is also the bisector of angle }A,$ [color=beige]. . [/color]$\text{prove: }AB\:=\:AC.$
Code:
              A
*
/|\
/1|2\
/  |  \
/   |   \
/    |    \
/     |     \
/     3|4     \
B * - - - * - - - *C
D

$\text{Since }AD\text{ bisects }\angle A:\;\;\angle 1\=\angle 2$

$\text{Of course, }AD \,=\,AD.$

$\text{Since }AD \,\perp\,BC:\;\;\angle 3\= \angle 4\ = 90^o$

$\text{Hence: }\:\Delta ADB\: \begin{array}{c}\sim \\= \end{array} \:\Delta ADC\;\;\text{(a.s.a.)}$

$\therefore\;\;AB \,=\,AC$

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