Question

Measure of a median of an isosceles triangle is half of its base. Find the interior angles of the triangle.

• A. ${45}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. Insufficient data.

Answer 1

Answer: (A), (C)

${90}^o$

We know: Median of an isosceles triangle is a perpendicular bisector that divides the side into two equal parts.

Here, AD is the perpendicualr bisector.

For the $△ABD,$ AD = BD
$\therefore$ This is an isosceles triangle.
Hence, $\angle B=\angle y$
Also, $\angle ADB={90}^o$

$\therefore \angle B+\angle y+{90}^o={180}^o$
$\Rightarrow 2\angle y={180}^o-{90}^o={90}^o$
$\Rightarrow \angle y=\frac{{90}^o}{2}={45}^o$
$\Rightarrow \angle B=\angle y={45}^o$

Similarly, for the triangle $△ADC,$
$\angle DAC={45}^o$ and $\angle C={45}^o$

$\therefore \angle A=\angle BAD+\angle DAC={90}^o$

Hence, the interior angles are ${45}^o,{45}^o$ and ${90}^o.$