Problem — A right triangle with one angle double the other

Text

Let $ABC$ be a right triangle with the right angle at the vertex $C$, in which one of the acute angles is double the other. Prove that the hypotenuse is twice as long as the shorter leg.

Solution

Let $\widehat{B}$ denote the smaller acute angle and $\widehat{A}$ the larger one, with $\widehat{A} = 2\,\widehat{B}$. Since $\widehat{A} + \widehat{B} = 90^\circ$, we have $2\,\widehat{B} + \widehat{B} = 90^\circ$, hence $\widehat{B} = 30^\circ, \qquad \widehat{A} = 60^\circ.$ The smaller acute angle, $\widehat{B} = 30^\circ$, is opposite the shorter leg $AC$; the hypotenuse is $AB$.

Reflect the triangle across the line $BC$ (the longer leg). The vertex $A$ maps to the point $A'$; since $AC \perp BC$ and $C$ lies on $BC$, the point $C$ is the midpoint of the segment $AA'$, hence $AA' = 2\,AC.$ In triangle $ABA'$ the reflection preserves lengths and angle measures, so $BA' \cong BA$ and $\widehat{A'BC} \cong \widehat{ABC} = 30^\circ$. It follows that $\widehat{A'BA} = \widehat{A'BC} + \widehat{ABC} = 30^\circ + 30^\circ = 60^\circ.$ Triangle $ABA'$ is isosceles (because $BA \cong BA'$) and has apex angle $\widehat{A'BA} = 60^\circ$; therefore its two base angles each measure $\dfrac{180^\circ - 60^\circ}{2} = 60^\circ$. Having all angles equal to $60^\circ$, triangle $ABA'$ is equilateral, so $AA' \cong AB.$ Combining the two relations, $AB = AA' = 2\,AC$: the hypotenuse is twice the shorter leg. $\boxed{AB = 2\,AC}$