Question

In a triangle $ABC$, right-angled at $B$, $BD$ is drawn perpendicular to $AC$. Such that:$\angle CBD=\angle A$

State true or false.

• A. True
• B. False

Answer 1

The correct option is A True

Given, $\angle ABC={90}^{\circ }$ and $\angle CDB={90}^{\circ }$

In $△ABC$,

$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$

$\angle BAC+{90}^{\circ }+\angle ACB={180}^{\circ }$

$\angle BAC+\angle ACB={90}^{\circ }$...(I)
In $△CBD$,

$\angle CBD+\angle CDB+\angle BCD={180}^{\circ }$

$\angle CBD+{90}^{\circ }+\angle BCD={180}^{\circ }$

$\angle CBD+\angle BCD={90}^{\circ }$...(II)

Equating I and II

$\angle BAC+\angle ACB=\angle CBD+\angle BCD$

$\angle BAC=\angle CBD$ (Since, $\angle ACB=\angle BCD$)