Question

$△ABC$ is a right triangle at $\angle B$ such that $\angle BCA=2\angle BAC.$ Show that $AC=2BC.$

Answer 1

Solution:
Given: $△ABC$ is a right triangle at $\angle B$ such that
$\angle BCA=2\angle BAC$
Construction: Produce $CB$ to a point $D$ such that $BC=BD$ and join $AD.$



In $△ABC$ and $△ABD,$
$BC=BD\text{(By construction)}$
$AB=AB\text{(Common)}$
$\angle ABC=\angle ABD\left({90}^{\circ }\text{each}\right)$
$\therefore △ABC\cong △ABD\text{(by SAS axiom)}$

$\Rightarrow \angle CAB=\angle DAB$ and
$AC=AD\dots \left(i\right)\text{(by C.P.C.T)}$
Let $\angle CAB=\angle DAB=x$
Thus, $\angle CAD=\angle CAB+\angle BAD=x+x=2x\dots \left(ii\right)$
$\text{Since},AC=AD$
$\Rightarrow \angle BCA=\angle BDA$
Also, $\angle BCA=2\angle BAC\text{(given)}$
$=2x$
$\Rightarrow \angle BDA=\angle BCA=2x\dots \left(iii\right)$
$\Rightarrow \angle ACD=\angle ADC=\angle CAD$ $\text{[From (ii) and (iii)]}$
$\therefore △ACD\text{is an equilateral triangle.}$
$\Rightarrow AC=CD$
$\Rightarrow AC=BC+BD$
$\Rightarrow AC=2BC(\text{Since}BC=BD)$

Hence, proved.