Question

In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate $\angle CDE$

• A. ${52}^o$
• B. ${54}^o$
• C. ${62}^o$
• D. none of the above

Answer 1

The correct option is A ${52}^o$

Given, $AB=AC$
$\angle ABC=\angle ACB=x$ (Isosceles triangle property)
$\angle BAF=\angle ABC+\angle ACB$
$128=x+x$
$2x=128$
$x={64}^{\circ }$
$\angle ABC=\angle ACB={64}^{\circ }$
GIven, $DE\parallel BC$,
$\angle ABC=\angle ADE=64$ (Corresponding angles)
In $△BDC$,
$BC=CD$ (Given)
Thus, $\angle BDC=\angle DBC={64}^{\circ }$
Now, $\angle ADE+\angle CDE+\angle BDC=180$
$64+\angle CDE+64=180$
$\angle CDE={52}^{\circ }$