Question

The bisector of the angle A of a triangle ABC meets BC in D, and BC is produced up to E. Such that:
$\angle ABC$+$\angle ACE$ = 2$\angle ADC$

• A. True
• B. False

Answer 1

The correct option is A True

In $△ABD$,
$\angle ADC=\angle ABC+\frac{1}{2}\angle BAC$ (Exterior angle property)
or $2\angle ADC=2\angle ABC+\angle BAC$
or $\angle BAC=2(\angle ADC-\angle ABC)$ ..(I)
In $△ABC$
$\angle ACE=\angle ABC+\angle BAC$ (Exterior angle property)
$\angle BAC=\angle ACE-\angle ABC$....(II)
Equating I and II,
$2(\angle ADC-\angle ABC)=\angle ACE-\angle ABC$
Hence, $\angle ACE+\angle ABC=2\angle ADC$