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
- Let △ABC be the right-angled triangle with ∠B=90∘ and ∠ACB=2∠CAB.
Let ∠CAB be equal to x∘. So, ∠ACB=2×∠CAB=2×x∘=2x∘. - We see that AC is the hypotenuse of △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 △ABC and △ABD, we have ∠ABC=∠ABD=90∘[∠ABD=90∘by linear pair]AB=AB[Common]BC=BD[By construction]∴
- 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} .