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

$AB$ divides $\angle DAC$ in the ratio $1 : 3$ and $AB = DB$.

To do:

We have to determine the value of $x$.

Solution:

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

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

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

$=72^{\circ}$

$\angle \mathrm{DAB}=72^{\circ} \times\frac{1}{1+3}$

$=\frac{72^{\circ} \times 1}{4}$

$=18^{\circ}$

$\angle \mathrm{CAB}=72^{\circ}\times\frac{3}{4}$

$=54^{\circ}$

Angles opposite to equal sides are equal.

This implies,

$\angle \mathrm{DAB}=\angle \mathrm{ADB}=18^{\circ}$            (Since $AB = DB$)

In $\triangle \mathrm{ADC}$,

$\angle \mathrm{CAE}=\angle \mathrm{BDA}+\angle \mathrm{ACD}$

$108^{\circ}=18^{\circ}+x$

$x=108^{\circ}-18^{\circ}$

$x=90^{\circ}$

The value of $x$ is $90^o$.

Tutorialspoint

Updated on: 10-Oct-2022

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