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Let $\bigtriangleup ABC$ has$ \angle A= \frac { \pi } {2} $.$ACC_1 A_2 $ and $BAA_1 B_1 $ both are external squares to $\bigtriangleup ABC $.$AB \cap B_1C=D_1$ and $AC \cap BC_1 =D_3$.The internal angle bisector of $ \angle BAC $ meets $BC$ at $D_2$.Prove that $AD_1D_2D_3$ is a square.

It's my one of the most favorite self-made problems.I've made its 6 different proofs.It's easy to prove it by calculation like complex,co-ordinate or by straight-cut calculation.But it can be done using only the figure!!!So I hope everyone will try to find such nice proofs.

$BB_1,CC_1$ are parallel to $A_1C,BA_2$ respectively.
now,$\bigtriangleup B_1D_1B\sim \bigtriangleup AD_1C$
then,$\frac{B_1B}{AC}=\frac{BD_1}{AD_1}$
$\Rightarrow \frac{AB}{AC}=\frac{BD_1}{AD_1}$
$\Delta CC_1D_3\sim \Delta ABD_3$
directly,$\frac{AC}{AB}=\frac{CD_3}{AD_3}..............(1)$
last 2 equation,$\frac{BD_1}{AD_1}=\frac{AD_3}{CD_3}$
$\Rightarrow \frac{AB}{AD_1}=\frac{AC}{CD_3}$ [addding 1]
$\Rightarrow \frac{AB}{AC}=\frac{AD_1}{CD_3}$
comparing with $(1)$, $AD_1=AD_3$
$AD_2$ bisects $\angle A$.
$\frac{AB}{AC}=\frac{BD_2}{CD_2}$
$\Rightarrow \frac{BD_2}{CD_2}=\frac{AD_1}{CD_3}$
$AD_1,D_2D_3$ are parallel.this info and $AD_1=AD_3$ enough to show $AD_1D_2D_3$ is a square.

I found out the same proof as photon.
And I know nothing but euclidean yet.