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,∠EAC = 108° and DB = AB

Since,DB = AB 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,

∠EAC + ∠DAC =- 180°

∠DAC + 180° = 180°

∠DAC = 180° - 108°

∠DAC = 72°

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

So, let

\[\angle DAB = y, \angle BAC = 3y\]

Thus, 

\[y + 3y = \angle DAC\]

\[ \Rightarrow 4y = 72^\circ \]

\[ \Rightarrow y = \frac{72^\circ}{4}\]

\[ \Rightarrow y = 18^\circ\]

Hence.

\[\angle DAB = 18^\circ, \angle BAC = 3 \times 18^\circ = 54^\circ\]

Using (1)

∠BDA = ∠DAB

∠BDA = 18°

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

\[\angle EAC = \angle ADC + x\]

\[ \Rightarrow 108^\circ = 18^\circ + x\]

\[ \Rightarrow x = 90^\circ\]