Question

State TRUE or FALSE

In a triangle ABC, right-angled at B, BD is drawn perpendicular to AC. hence, $\angle$CDB = $\angle$A.

• A. True
• B. False

Answer 1

The correct option is B False

Given, $\angle ABC={90}^o$ and $\angle CDB={90}^o$
In $△ABC$,
$\angle BAC+\angle ABC+\angle ACB={180}^o$
$\angle BAC+\angle ACB+90={180}^o$
$\angle BAC+\angle ACB={90}^o$...(I)
In $△CBD$,
$\angle CBD+\angle CDB+\angle BCD={180}^o$
$\angle CBD+\angle BCD+{90}^o={180}^o$
$\angle CBD+\angle BCD={90}^o$...(II)
Equating I and II
$\angle BAC+\angle ACB=\angle CBD+\angle CDB$
$\angle BAC=\angle CBD$ ........... (Since, $\angle ACB=\angle BCD$)