Question

In figure, $\angle A={35}^0,\angle ABC={100}^0$ and $BD\perp AC$, prove that $\Delta BDC$ is an isosceles triangle.

Answer 1

In $\Delta ABC$,
$\angle A+\angle B+\angle C={180}^0$ (the sum of the interior angles of a $\Delta$ is equal to ${180}^0$)
${35}^0+{100}^0+\angle C={180}^0$
$\angle C={180}^0–{135}^0$
$\angle C={45}^0$
But in $\Delta ABD,BD\perp AC$
i.e., $\angle ADB={90}^0$
and $\angle A={35}^0$ (given)
$\angle ABD={180}^0–{125}^0={55}^0$
$\angle DBC={100}^0–{55}^0={45}^0$
Now, $\angle DCB=\angle DBC={45}^0$ (Proved above)
Hence $BDC$ is an isosceles triangle.