AD is an altitude of an isosceles triangle ABC in which AB = AC. Which of the following is true?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → BD = CD
In an isosceles triangle, the altitude to the base also bisects the base.
In right triangles ABD and ACD: AB \(=\) AC (given), AD is common, and \(\angle ADB=\angle ADC=90^\circ\)
So \(\triangle ABD\cong\triangle ACD\) (RHS)
Hence BD \(=\) CD
