### If one acute angle is double than the other in a right-angled triangle, prove that the hypotenuse is double the smallest side.

$AC = 2BC$

Step by Step Explanation:
1. Let $\triangle ABC$ be the right-angled triangle with $\angle B = 90^ \circ$ and $\angle ACB = 2 \angle CAB$.

Let $\angle CAB$ be equal to $x^ \circ$. So, $\angle ACB = 2 \times \angle CAB = 2 \times x^ \circ = 2x^ \circ.$
2. We see that $AC$ is the hypotenuse of $\triangle ABC$.

Also, the side opposite to the smallest angle is the smallest. Thus, $AC$ is the smallest side.

Now, we need to prove $AC = 2BC$.
3. Let us extend $CB$ to $D$ such that $CB = BD$ and join point $A$ to point $D$.
4. In $\triangle ABC$ and $\triangle ABD$, we have \begin{aligned} & \angle ABC = \angle ABD = 90^ \circ && [\angle ABD = 90^ \circ \text{by linear pair]}\\ & AB = AB && \text{[Common]} \\ & BC = BD && \text{[By construction]} \\ \therefore {\space} & \triangle ABC \cong \triangle ABD && \text{[By SAS-criterion]} \end{aligned}
5. As corresponding parts of congruent triangles are equal, we have \begin{aligned} AC = AD \text{ and } \angle DAB = \angle CAB = x^ \circ \end{aligned} $$\therefore {\space} \angle DAC = \angle DAB + \angle CAB = x^ \circ + x^ \circ = 2x^ \circ$$
6. Now, in $\triangle ACD$, we have \begin{aligned} & \angle DAC = \angle ACD = 2x^ \circ \\ \implies & AD = CD && \text{[Sides opposite to equal angles are equal.]}\\ \implies & AC = CD && [As \space AD = AC] \\ \implies & AC = 2BC && [As \space CD = 2BC] \end{aligned}
7. $\text{Hence, } \bf {AC = 2BC}$. 