Question

Find the measure of the angle $\angle ADC$, if $AE=AD$ and is an equilateral triangle $\Delta ABC$ and $BCDE$ is a square.

• A. ${45}^o$
• B. ${30}^o$
• C. ${15}^o$
• D. ${18}^o$
• E. None of these

Answer 1

The correct option is C ${15}^o$

REF. Image

ABC is an equilateral $△$.

$\Rightarrow AB=BC=CD$ _____ (I)

BCDE is a square

$\Rightarrow BC=CD=DE=EB$ ____ (II)

From (I) & (II)

$AB=BC=CD=DE=EB$

$=CD$

$\because AC=CD$

$\Rightarrow △ACD$ is an isosceles triangle

We know that angle opposite to equal

sides of an isosceles triangle are equal

$\Rightarrow \angle ADC=\angle CAD$

Sum of angles of a triangle = ${180}^{\circ }$

$\angle ADC+\angle CAD+\angle ACD={180}^{\circ }$

$\angle ADC+\angle ADC+{150}^{\circ }={180}^{\circ }$ $(\angle ACD=\angle ACB+\angle BCD)$

$2\angle ADC={30}^{\circ }$

$\angle ADC={15}^{\circ }$

$\therefore \angle ADC={15}^{\circ }$