2 squares drawn outside of triangle, isosceles triangle criterion

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

2 squares drawn outside of triangle, isosceles triangle criterion

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Mathematical Excalibur P48 (v2)

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30630 posts

#1 h Aug 10, 2018, 11:29 PM • 2 Y

Y by Adventure10, Mango247

Squares $ABDE$ and $BCFG$ are drawn outside of triangle $ABC$ . Prove that triangle $ABC$ is isosceles if DG is parallel to $AC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

43 posts

#2 h Aug 11, 2018, 12:05 AM • 1 Y

Y by Adventure10

Let $B$ be the origin of the complex plane and let $a,c$ be the complex affixes of $A,C$ . Assume $ABC$ is labeled counterclockwise.

Clearly $d = ia, g = -ic$ . The parralel condition translates to
$i\frac{a+c}{a-c}= -i\frac{\bar{a} + \bar{c}}{\bar{a} - \bar{c}}.$
If you denote $k=\frac{a}{c}$ , the equality becomes
$\frac{k+1}{k-1} + \frac{\bar{k} + 1}{\bar{k}-1} = 0,$ which reduces to $|k|^2 = 1,$ therefore $|a| = |c|$ and so $ABC$ is isosceles.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a