Question

In the figure, ABC is an isosceles triangle in which AB = AC. If D and E are the mid-points of sides AB and BC respectively and $\angle$ DOE = 120º, find
$\angle$ ODE.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is A

${30}^{\circ }$

In $△BDC$ and $△CEB$

$DB=CE(AB=AC;\frac{AB}{2}=\frac{AC}{2}$ )

$\angle DBC=\angle ECB\left(ABCisanisoscelestriangle\right)$

$BC=CB\left(Common\right)$

By SAS congruence condition, $△BDC\cong △CEB$

So, $\angle BDC=\angle BEC\left(Correspondingpartofcongruenttriangles\right)$

In ADE, AD = AE. So,
$\angle ADE=\angle AED$
This means that $\angle BDE=\angle CED$

Since
$\angle BDE=\angle CEDand\angle BDC=\angle BEC,\angle ODE=\angle OED=x\left(Let\right)$

In $△ODE,x+x+\angle DOE={180}^{\circ }$ (Angle sum property of a triangle)

$2x+{120}^{\circ }={180}^{\circ }$

$x={30}^{\circ }=\angle ODE$