Question

In a triangle ABC, the bisectors of angles ABC and ACB meet at M.
hence,
$\angle BMA={90}^{\circ }+\frac{1}{2}\angle A$.

State true or false.

• A. True
• B. False

Answer 1

The correct option is B False

In $△BMC$,
Sum of all angles of a triangle $={180}^o$
$\angle BMC+\angle MBC+\angle MCB={180}^o$
$\angle BMC={180}^o-(\angle MBC+\angle MCB)$
$\angle BMC={180}^o-\frac{1}{2}(\angle B+\angle C)$ (Since, MB and MC bisect $\angle B$ and $\angle C$)
$\angle BMC={180}^o-\frac{1}{2}(180-\angle A)$ (Using angle sum property to triangle ABC)
$\angle BMC={180}^o-{90}^o+\frac{1}{2}\angle A$
$\therefore \angle BMC={90}^o+\frac{1}{2}\angle A$