Question

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Answer 1

Step-by-step explanation:

Correct option is

C

Both A and B

A)

Altitude AD bisects ∠BAC

In △BAD and △CAD

AD=AD (Common Side)

∠ADB=∠ADC=90

o

(AD is altitude)

∠BAD=∠CAD (AD bisects ∠BAC)

∴△BAD≅△CAD(A.S.A)

⇒AB=AC(C.P.C.T.C.)

Thus, △BAC is an isosceles triangle.

B)

Median AD is perpendicular to BC

In △BAD and △CAD

AD=AD (Common Side)

∠ADB=∠ADC=90

o

(AD is perpendicular to BC)

BD=CD (AD is Median)

∴△BAD≅△CAD(S.A.S.)

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⇒AB=AC(C.P.C.T.C.)

Thus, △BAC is an isosceles triangle.

solution

Answer 2

Answer:

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