Question

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

Answer 1

In the given figure,and

Since,and angles opposite to equal sides are equal. We get,

$\angle BDA=\angle BAD.....\left(1\right)$

Also, EAD is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Further, it is given AB divides in the ratio 1 : 3.

So, let

$\angle DAB=y,\angle BAC=3y$

Thus,

$$\begin{gathered} y+3y=\angle DAC \\ \Rightarrow 4y=72° \\ \Rightarrow y=\frac{72°}{4} \\ \Rightarrow y=18° \end{gathered}$$

Hence, $\angle DAB=18°,\angle BAC=3\times 18°=54°$

Using (1)

Now, in ΔABC , using the property, “exterior angle of a triangle is equal to the sum of its two opposite interior angles”, we get,

$$\begin{gathered} \angle EAC=\angle ADC+x \\ \Rightarrow 108°=18°+x \\ \Rightarrow x=90° \end{gathered}$$

Therefore,.