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 $\angle DOE={120}^{\circ },$ find $\angle ODE$

• A. 30º
• B. 45º
• C. 60º
• D. 75º

Answer 1

The correct option is A

30º

In $△BDC$ and $△CEB$

$DB=CE$ [$\because AB=AC;\frac{AB}{2}=\frac{AC}{2}$]

$\angle DBC=\angle ECB$ [$\because ABC$ is an isosceles triangle]

$BC=BC$ [Common]

So, by $SAS$ congruence condition, $△BDC\cong △CEB$

Hence, $\angle BDC=\angle BEC$ [Corresponding of congruent triangles]

In $△ADE,AD=AE.$ So, $\angle ADE=\angle AED.$

This means that $\angle BDE=\angle CED$

Since, $\angle BDE=\angle CED$ and $\angle BDC=\angle CEB,$

Let $\angle ODE=\angle OED$

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

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

$\Rightarrow 2x=180-120$

$\Rightarrow x={30}^{\circ }$

Hence, $\angle ODE={30}^{\circ }$