Question

In the adjoining figure, find the measure of $\angle BAC$, if $\angle ABD=\angle CAD$ and $\angle BAD=\angle ACD$.

• A. ${120}^o$
• B. ${60}^o$
• C. ${75}^o$
• D. ${90}^o$

Answer 1

Answer: (D)

${90}^o$

Given that:

$\angle ABD=\angle CAD$ and $\angle BAD=\angle ACD$

In $△ABC$, we have

$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$

$\angle ABD+\angle DCA+\angle CAB={180}^{\circ }$

$\angle CAD+\angle BAD+\angle CAB={180}^{\circ }$

$\angle CAB+\angle CAB={180}^{\circ }$

$2\angle CAB={180}^{\circ }$

$\therefore \angle CAB={90}^{\circ }$