Question -
Answer -
In the given figure,
and
Since,and angles opposite to equal sidesare equal. We get,
∠BDA=∠BAD .....(1)∠BDA=∠BAD .....1
Also, EAD is a straight line. So, using theproperty, “the angles forming a linear pair are supplementary”, we get,
Further, it is given AB divides
in the ratio 1 : 3.
So, let
∠DAB=y, ∠BAC=3y∠DAB=y, ∠BAC=3y
Thus,
y+3y=∠DAC⇒4y=72°⇒y=72°4⇒y=18°y+3y=∠DAC⇒4y=72°⇒y=72°4⇒y=18°
Hence, ∠DAB=18°, ∠BAC=3×18°=54°∠DAB=18°, ∠BAC=3×18°=54°
Using (1)
Now, in ΔABC , using the property, “exterior angleof a triangle is equal to the sum of its two opposite interior angles”, we get,
∠EAC=∠ADC+x⇒108°=18°+x⇒x=90°∠EAC=∠ADC+x⇒108°=18°+x⇒x=90°
Therefore,