## Solution.

From the condition that $\triangle A B D \equiv \triangle C A E$, it follows that $\angle D A B \equiv \angle E C A$. Let $m(\angle D A B)=\alpha=m(\angle E C A)$. Then $m(\angle A D B)=90^{\circ}-\alpha=m(\angle A E C)$. Let $A B \cap E C=\{F\}$. In triangle $A F E$, with $m(\angle E A F)=\alpha$ and $m(\angle A E F)=90^{\circ}-\alpha$, we obtain $m(\angle E F A)=90^{\circ}$. Thus, $C E \perp A B$.

From the condition that $\triangle A B D \equiv \triangle C A E$, it follows that $A B \equiv C A, D B \equiv E A$. Since $A B \equiv C A$, it follows that $\triangle A B C$ is isosceles. Thus, $m(\angle A B C)=m(\angle B C A)=0.5 \cdot\left(180^{\circ}-m(\angle B A C)\right)=45^{\circ}+\frac{\alpha}{2}$.

Since $E B \perp B C$, we obtain $m(\angle E B F)=90^{\circ}-m(\angle A B C)=45^{\circ}-\frac{\alpha}{2}$.

Since $A B \perp B D$, we obtain $m(\angle E B D)=90^{\circ}-m(\angle E B F)=45^{\circ}+\frac{\alpha}{2}$.

Since $\triangle E F B$ is a right triangle, with $m(\angle E B F)=45^{\circ}-\frac{\alpha}{2}$, we obtain $m(\angle B E F)=45^{\circ}+\frac{\alpha}{2}$.

Since $\triangle A F E$ is a right triangle, with $m(\angle E A F)=\alpha$, we obtain $m(\angle A E F)=90^{\circ}-\alpha$.

Since $m(\angle D E B)=180^{\circ}-m(\angle B E F)-m(\angle A E F)=45^{\circ}+\frac{\alpha}{2}$, we obtain $\angle D E B \equiv \angle D B E$. Thus, $\triangle D E B$ is isosceles with $D E \equiv D B$. Since $D B \equiv E A$ and $D E \equiv D B$, we obtain $D E \equiv E A$. From here, we obtain $E$ - the midpoint of $A D$. Thus, $B E$ is the median drawn

![](https://cdn.mathpix.com/cropped/2024_06_06_50700f3d6002395bf0ceg-2.jpg?height=631&width=528&top_left_y=850&top_left_x=1384)
to the hypotenuse $D A$ in the right triangle $D B A$. Since $B E$ is the median drawn to the hypotenuse $D A$, we obtain $B E=0.5 D A=D E$. Therefore, $\triangle D E B$ is equilateral.

Since $\triangle D E B$ is equilateral, it follows that $m(\angle E D B)=60^{\circ}=90^{\circ}-\alpha$, or $\alpha=30^{\circ}$. Therefore, $m(\angle B A C)=$ $60^{\circ}$.

Since $\triangle A B C$ is isosceles with $m(\angle B A C)=60^{\circ}$, it follows that $\triangle A B C$ is equilateral with side lengths equal to $10 \mathrm{~cm}$. Therefore, $P_{\triangle A B C}=30 \mathrm{~cm}$.

Answer: $P_{\triangle A B C}=30 \mathrm{~cm}$.
7.3. Let $A B C$ be an acute-angled triangle. The perpendicular from $A$ to $A C$ and the perpendicular from $B$ to $A B$ intersect at point $D$. On the side $A D$, an interior point $E$ is taken such that triangles $A B D$ and $C A E$ are congruent. The length of side $A B$ is $10 \mathrm{~cm}$, and $E B \perp B C$. Calculate the perimeter of triangle $A B C$.