Question

In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side. [4 MARKS]

Answer 1

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

Given: A $\Delta$ABC in which $\angle$B = ${90}^{\circ }$and $\angle$ACB = 2 $\angle$CAB.

To prove: AC = 2 BC.

Construction: Produce CB to D such that BD = CB. Join AD.

Proof: In $\Delta$s ABD and ABC, we have
BD = BC [By construction]
AB = AB [Common side]
and, $\angle$ABD = $\angle$ABC [Each equal to ${90}^{\circ }$]
So,
$\Delta$ ABD $\cong$ $\Delta$ ABC [SAS criterion of congruence]

$\Rightarrow$ AD = AC [C.P.C.T.C]

In $\Delta$ ACD, AD = AC
Since $\angle$ACB = ${60}^{\circ }$, $\angle$ADB also equal to ${60}^{\circ }$

Hence, the $\Delta$ ACD is an equilateral triangle.

$\Rightarrow$ AD = AC = CD

$\Rightarrow$ AC = CD

$\Rightarrow$ AC = 2BC

$\Rightarrow$ Hypotenuse is twice the smallest angle.