Question

State true or false:

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

• A. True
• B. False

Answer 1

The correct option is A True

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