Question

AD is an altitude of an isosceles triangle ABC is which AB=AC .prove that angle BAD = angleDAC. SOLVE FAST PLZZ

Answer 1

Step-by-step explanation:

In tringle ABD & ACD

AB=AC (GIVEN)

ANGLE ADB=ANGLE ADC(BOTH ARE RIGHT ANGLES)

AD=AD(COMMON)

SO BOTH THE TRIANGLES ARE CONGRUENT

THEREFORE BD=CD(CPCT)

ANGLE BAD=ANGLE CAD(CPCT)

HENCE AD BISECTS BOTH BC AND ANGLE A

Answer 2

Given:

AD is the altitude of an isosceles triangle ABC, in which AB = AC.

To Find:

prove that ∠BAD = ∠DAC

Solution:

It is given that

AB = AC

Since AD is the altitude,

∠BDA = ∠ADC = 90°

Side AD is common to both the sides of ∆ABD and ∆ADC, so

AD = AD    (common)

∴ ∆ABD ≅ ∆ADC

By Corresponding Part of Congruent Triangles, we get

∠BAD =∠DAC

Hence, ∠BAD =∠DAC.