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

**Answer:**

^@ AC = 2BC ^@

**Step by Step Explanation:**

- 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.^@ - 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^@. - Let us extend ^@CB ^@ to ^@ D ^@ such that ^@ CB = BD ^@ and join point ^@ A ^@ to point ^@ D ^@.
- 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} @^
- 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 @^
- 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}@^
- ^@\text{Hence, } \bf {AC = 2BC} ^@.