Question

In the figure, $AB$ divides $\angle DAC$ in the ratio $1:3$ and $AB=DB$. Determine the value of $x.$

Answer 1

In the figure, $AB=DB$

$AB$ divides $\angle DAC$ in the ratio $1:3$ and $\angle CAE={108}^{\circ }$

But $\angle CAE+\angle DAC={180}^{\circ }$ (Linear pair )

$\Rightarrow {108}^{\circ }+\angle DAC={180}^{\circ }$

$\Rightarrow \angle DAC={180}^{\circ }-{108}^{\circ }={72}^{\circ }$

$\therefore$ $AB$ divides $\angle DAC$ in the ratio $1:3$

$\therefore \angle DAB={72}^{\circ }\times \left(\frac{1}{1+3}\right)={72}^{\circ }\times \left(\frac{1}{4}\right)={18}^{\circ }$

and $\angle CAB={72}^{\circ }\times \frac{3}{4}={54}^{\circ }$
$\because$ $AB=BD$
$\angle DAB=\angle ADB={18}^{\circ }$ [Angles opposite to equal sides are equal]
Now in $\Delta ADC,$
Ext angle .$\angle CAE=\angle BDA+\angle ACD$ [$\because$ Exterior angle is equal to sum of opposite non-adjacent interior angles]
$\Rightarrow {108}^{\circ }={18}^{\circ }+x$
$\Rightarrow x={108}^{\circ }-{18}^{\circ }={90}^{\circ }$
$\therefore x={90}^{\circ }$