Question

AD bisects angle A of triangle ABC, where D lies on BC and angle C is greater than angle B. then angle ADB is greater than angle ADC.

• A. True
• B. False

Answer 1

The correct option is A True

In $△ADB$
Sum of all angles of a triangle $={180}^o$
$\therefore \angle B+\angle ADB+\frac{1}{2}\angle A=180$
$\angle ADB={180}^o-\angle B-\frac{1}{2}\angle A$
Similarly, $\angle ADC={180}^o-\angle C-\frac{1}{2}\angle A$
Since, $\angle C>\angle B$
Hence, $\angle ADB>\angle ADC$