Question

Sides $AB$ and $AC$ of a triangle $ABC$ are equal. $BC$ is produced through $C$ upto point $D$ such that $AC=CD$. $D$ and $A$ are joined and produced (through vertex $A$) upto point $E$. If angle $BAE={108}^{\circ }$;

The value of angle $ADB$ is ${48}^o$

• A. True
• B. False

Answer 1

The correct option is B False

Given, $AB=AC=CD$, $\angle BAE={108}^{\circ }$
In $△ABC$, $AB=AC$
Hence, by Isosceles triangle property, $\angle ABC=\angle ACB=x$
$\angle BAE=\angle ABC+\angle ACB$ (Exterior angle property)
Therefore, $108=x+x$
$\Rightarrow 2x=108$
$\Rightarrow x={54}^{\circ }$
Hence, $\angle ABC=\angle ACB={54}^{\circ }$
Now, In $△CAD$, $AC=CD$
Hence, $\angle CAD=\angle CDA=y$
$\angle ACB=\angle CAD+\angle CDA$ (Exterior angle property)
Therefore, $54=y+y$
$\Rightarrow 2y=54$
$\Rightarrow y=27$
Thus $\angle ADB={27}^{\circ }$