In right triangle $A B C, A X=A D$ and $C Y=C D$, as shown. The measure of $\angle X D Y$ is
(A) $35^{\circ}$
(B) $40^{\circ}$
(C) $45^{\circ}$
(D) $50^{\circ}$
(E) not determined by this information

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-134.jpg?height=366&width=526&top_left_y=2137&top_left_x=1271)
In right triangle $A B C, A X=A D$ and $C Y=C D$, as shown. The measure of $\angle X D Y$ is
(A) $35^{\circ}$
(B) $40^{\circ}$
(C) $45^{\circ}$
(D) $50^{\circ}$
(E) not determined by this information

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-263.jpg?height=415&width=588&top_left_y=1462&top_left_x=1316)

## Solution

Let $\angle D Y C=\theta$. Then $\angle Y D C=\theta$ ( $\triangle Y D C$ is isosceles).

Then $\angle Y C D=180^{\circ}-2 \theta$ (sum of angles in $\triangle Y D C$ ).

Then $\angle C A B=2 \theta-90^{\circ}$ (sum of angles in $\triangle A B C$ ).

In $\triangle A X D, \angle A X D=\angle A D X$ so $2 \theta-90^{\circ}+2 \angle A X D=180^{\circ}$.

So $\angle A X D=135^{\circ}-\theta$ and $\angle A D X=135-\theta$.

But $\angle A D X+\angle X D Y+\angle Y D C=180^{\circ}$.

Therefore, $135-\theta+\angle X D Y+\theta=180^{\circ}$.

Thus, $\angle X D Y=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-263.jpg?height=412&width=588&top_left_y=1976&top_left_x=1316)

ANSWER: (C)