Another Proof of the Hypotenuse Leg Theorem

Given two triangles, if their hypotenuse are congruent, and one pair of their legs are congruent then the two triangles are congruent. In this post, we are going to prove this theorem.

In the figure below, ABC and DEF are right triangles with right angles at C and F, respectively.

hypotenuse leg 1

It is given that $latex \overline{AB} \cong \overline{DE}$ and $latex \overline{AC} \cong \overline{DF}$. We are going to prove that $latex \triangle ABC \cong \triangle DEF$. 


In $latex \triangle  DEF$, construct ray $latex EF$ and mark $latex G$ so that $latex \overline{FG} \cong \overline{BC}$ (S). Since it is given $latex \overline{DF} \cong \overline{AC}$ (S) and $latex \angle DFA \cong ACB$ (A), then by SAS Congruence, $latex \triangle ABC \cong \triangle DGF$. Since corresponding parts of congruent triangles are congruent, $latex \overline{AB} \cong \overline{DG}$.

hypotenuse leg 2

It is given that $latex \overline{AB} \cong \overline{DE}$, so $latex \overline{DE} \cong \overline{DG}$ by the transitive property of congruence. This means that $latex \triangle DGE$ is an isosceles triangle. Since the angles opposite of the congruent sides of a triangle are congruent, $latex \angle G \cong \angle E$. Therefore, $latex \triangle DFG \cong \triangle DFE$ by AAS Congruence. Therefore, $latex \triangle ABC \cong \triangle DEF$ by the transitive property of congruence.

Leave a Reply

Your email address will not be published. Required fields are marked *