Problem 7.6. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively. It is known that $A B=B D, \angle A B D=46^{\circ}, \angle D E C=90^{\circ}$. Find $\angle B D E$, if it is known that $2 D E=A D$.
Answer: $67^{\circ}$.

Solution. Draw the height $B M$ in triangle $A B D$. Since this triangle is isosceles, $M$ bisects $A D$ (Fig. 7.6). This means that $M D = A D / 2 = D E$.

Then the right triangles $B D M$ and $B D E$ are congruent by the leg ( $D M = D E$ ) and the hypotenuse ( $B D$ - common). From this, it is not difficult to find the required angle:

$$
\angle B D E = \angle B D A = \frac{180^{\circ} - \angle A B D}{2} = \frac{180^{\circ} - 46^{\circ}}{2} = 67^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c7c94bf2f2dc4a178b54g-4.jpg?height=343&width=505&top_left_y=93&top_left_x=468)

Fig. 2: for the solution of problem 7.6