Question

IN a triangle ABC ,AD bisects angle A and angle C greater than angle B .PROVE that angle ADB GREATE
R THAN angle ADC.

Answer 1

Step-by-step explanation:

In △ADB

Sum of all angles of a triangle =180°

∴∠B+∠ADB+ 1/2∠A=180°

∠ADB=180°−∠B−1/2∠A

Similarly, ∠ADC=180°−∠C−1/2∠A

Since, ∠C>∠B

Hence, ∠ADB>∠ADC