Question

AD is an altitude of an isosceles triangle ABC in which AB=AC.show that
1.AD bisects BC
2.AD bisects angle A​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Since diagonal of isosceles triangle is perpendicular on the side then,

<ADC=<ADB=90°

Now,in triangle ADC and triangle ADB

<ADC=<ADB=90°

<B=<C[AB=AC angle opposite to equal sides are equal]

AD=AD[Common]

so,triangle ADC is congruent to triangle ADB

AD=DB[CPCT]

so,AD bisect AB

and,<BAD=<CAD[CPCT]

so,AD bisect <A

HENCE PROVED........