Question

ABC is an isosceles triangle such that AB = AC and AD is the median to base BC, then, $\angle BAD=$

• A. ${55}^{\circ }$
• B. ${70}^{\circ }$
• C. ${35}^{\circ }$
• D. ${110}^{\circ }$

Answer 1

The correct option is A

${55}^{\circ }$

$In\Delta ABC,AB=AC$ and AD is median to BC.

$\therefore BD=DC$

$In\Delta ADB,\angle D={90}^{\circ },\angle B={35}^{\circ }$

$But\angle B+BAD+\angle D={180}^{\circ }$

(Sum of angles of a triangle)

$\Rightarrow {35}^{\circ }+\angle BAD+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BAD+{125}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BAD={180}^{\circ }-{125}^{\circ }={55}^{\circ }$