Congruence of two right triangles by equal hypothenuses and heights

Question

Two triangles ($ABC$ and $MNP$) are right (angles $ACB$ and $MPN$ are $90$ degrees each), the hypothenuses $AB = MN$ and the heights to the hypothenuses $CD = PQ$. Prove that the two triangles are congruent.

Answer 1

Let $E$ be the midpoint of $AB.$ Let $R$ be the midpoint of $MN.$

Show that $\triangle CDE \cong \triangle PQR$, then show that either $\angle CAB \cong \angle PMN$ or $\angle CAB \cong \angle PNM.$

Answer 2

You're given that the hypotenuses and the altitudes on them are congruent. Since the given segments determine the vertices of each triangle, they have determined both up to congruence, and the result follows.

Answer 3

Let $a$ and $b$ be the sides of ABC and $m$ and $n$ be the sides of MNP. Then, from the same hypotenuse and area, we have $$a^2+b^2=m^2+n^2$$ $$ab=mn$$

Two solutions due to symmetry: either $a=m,\>b=n$, or $a=n,\>b=m$, which yields congruent triangles either way.