Question

Sides $AB$ and $AC$ of a triangle $ABC$ are equal. $BC$ is produced through $C$ up to point $D$ such that $AC=CD.$ $D$ and $A$ are joined and produced (through vertex $A$ ) upto point $E.$ If $\angle BAE=108$ degrees . Find $\angle ADB.$

• A. ${27}^{\circ }$
• B. $3^{\circ }$
• C. $6^{\circ }$
• D. ${63}^{\circ }$

Answer 1

The correct option is A ${27}^{\circ }$

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)

$108=x+x$

$2x={108}^o$

$x={54}^{\circ }$

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)

${54}^o=y+y$

$2y={54}^o$

$y={27}^o$

thus, $\angle ADB={27}^{\circ }$