Question

In the given figure ABC is a right triangle and right angled at B such that $\angle BCA=2\angle BAC$.
Show that hypotenuse AC = 2BC.
(Hint : Produce CB to a point D that BC = BD)

Answer 1

In $△ABD$ and $△ABC$ we have $BD=BC$

$AB=AB$ [Common]

$\angle ABD=\angle ABC={90}^{\circ }$

$\therefore$ By SAS criterion of congruence we get

$△ABD\cong △ABC$

$\Rightarrow AD=AC$ and $\angle DAB=\angle CAB$ [By CPCT]

$\Rightarrow AD=AC$ and $\angle DAB=x$ [$\therefore \angle CAB=x$]

Now, $\angle DAC=\angle DAB+\angle CAB=x+x=2x$

$\therefore \angle DAC=\angle ACD$

$\Rightarrow DC=AD$ [Side Opposite to equal angles]

$\Rightarrow 2BC=AD$ since $DC=2BC$

$\Rightarrow 2BC=AC$ Since $AD=AC$

Hence proved.