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 AC respectively and $\angle DOE={120}^{\circ }$, then find $\angle ODE$.

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

Answer 1

The correct option is A

${30}^{\circ }$

Given, AB = AC.
$⟹\frac{AB}{2}=\frac{AC}{2}$
$⟹$ DB = CE $\cdots \left(i\right)$ (D and E are the midpoints of AB and AC respectively)

Consider $△BDC$ and $△CEB$.
DB = CE [from ($i$)]
$\angle DBC=\angle ECB$ ($△ABC$ is an isosceles triangle)
BC = CB (common)
$\therefore △BDC\cong △CEB$ (SAS congruency)
$⟹\angle BDC=\angle BEC$ --- (1) (CPCT)

In $△ADE$, AD = AE.
$⟹\angle ADE=\angle AED$ --- (2)
(In a triangle, angles opposite to equal sides are equal)

From (1) and (2), we must have,
$⟹\angle ODE=\angle OED=x$ (say)

In $△ODE$, $x+x+\angle DOE={180}^{\circ }$ (Angle sum property of a triangle)
$⟹2x+{120}^{\circ }={180}^{\circ }$
$⟹x=\angle ODE={30}^{\circ }$