Question

ΔABC is an isosceles triangle with AB=AC, AD is the altitude from A to side BC. Draw
a rough figure and prove that
(i) ΔADB ≅ ΔADC
(ii) ∠BAD = ∠CAD

Answer 1

Answer:

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

Step-by-step explanation: