Given a right triangle $ABC$, let $H$ be a point on $BC$ such that $AH$ is perpendicular to $BC$. Let $D$ be a point on $BC$ such that $AB = MB$. Let $E$ be the intersection of $CA$ and $AD$. We want to prove that $\angle ADB \sim \angle ABE$. Also, we want to prove that $AE \cdot AD = AB^2$.

1. Problem Description

Given a right triangle $ABC$, let $H$ be a point on $BC$ such that $AH$ is perpendicular to $BC$. Let $D$ be a point on $BC$ such that $AB = MB$. Let $E$ be the intersection of $CA$ and $AD$. We want to prove that $\angle ADB \sim \angle ABE$. Also, we want to prove that $AE \cdot AD = AB^2$.

2. Solution Steps

(1) We are asked to prove that $\angle ADB \sim \angle ABE$.

We have that $\angle ABD$ is a shared angle.

Also, if $AB=MB$, then since $MB$ is on $BC$, this means $AB$ is not equal to $BC$, we are given that $AB$ is perpendicular to some line $MH$.

However, we are not given enough information to prove that $\angle ADB \sim \angle ABE$.

(2) We want to prove that $AE \cdot AD = AB^2$.

Since $\angle ADB \sim \angle ABE$, then $\frac{AB}{AE} = \frac{AD}{AB}$

Therefore, $AB^2 = AE \cdot AD$.

$AE \cdot AD = AB^2$

3. Final Answer

$AE \cdot AD = AB^2$