Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?

![](https://cdn.mathpix.com/cropped/2024_05_06_d7ec6e60d14a8fdbf750g-30.jpg?height=476&width=517&top_left_y=91&top_left_x=468)
Answer: 17.

![](https://cdn.mathpix.com/cropped/2024_05_06_d7ec6e60d14a8fdbf750g-30.jpg?height=474&width=507&top_left_y=657&top_left_x=469)

Fig. 3: to the solution of problem 8.6

Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute-angled ($\angle ABD = \angle CBD < 90^\circ$, $\left.\angle BAD = \angle ADB = \frac{180^\circ - \angle ABD}{2} < 90^\circ\right)$, point $H$ lies on the segment $AB$.

Notice that the right triangles $BDH$ and $BDC$ are equal by the common hypotenuse $BD$ and the acute angle at vertex $B$. Therefore, $BH = BC$ and $DH = CD$.

Now, notice that the right triangles $ADH$ and $EDC$ are also equal by the hypotenuse $AD = ED$ and the leg $DH = CD$. Therefore, $EC = AH$.

Thus, $BD = BA = BH + AH = BC + EC = (7 + 5) + 5 = 17$.