$\triangle ABC$ is iscosceles and orthogonal, with $\angle BAC=90^o$, $DE//BC$ and $BD=BM$. Find how many degrees $\angle MDE$ is equal to

Question

$\triangle ABC$ is iscosceles and orthogonal, with $\angle BAC=90^o$, $DE//BC$ and $BD=BM$. Find how many degrees $\angle MDE$ is equal to.

I attempted to solve it by showing that $MBDE$ is a rhombus, but I didn't manage to prove that. Then I tried to find a parallelogram, but I couldn't find one anywhere. I think the answer is $\angle MDE=67.5^o$ although I'm not certain. Could you please show me how to solve the question?

Answer 1

$\bigtriangleup ABC\;$ is an isosceles right triangle, you can find all the angles of this triangle.

$\bigtriangleup BDM$ is an isosceles triangle, you can find all the angles of this triangle also.

Since $DE \parallel BC$, $\measuredangle ABC=\measuredangle ADE \;$ due to corresponding angles.

Now, $\measuredangle MDE \;$ can be found.