Question

Please tell the answer​

Answer 1

$\huge \sf\bf \underline \blue { \: \: \: answer✿༆ \: \: \: \: \: \: }$

Given:-

AD is the bisector of ∠A So we have, ∠DAB = ∠DAC

(1) AD ⊥BC

So we have, ∠BDA = ∠CDA = 90o

To prove,

ΔABC is isosceles. Proof, In ΔDAB and ΔDAC, ∠BDA = ∠CDA = 90o DA = DA (common) ∠DAB = ∠DAC (from 1) By ASA congruence property, ΔDAB ≅ ΔDAC AB = AC

Hence,

ΔABC is isosceles. !!!

Answer 2

Answer:

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Given:-

AD is the bisector of ∠A So we have, ∠DAB = ∠DAC

(1) AD ⊥BC

So we have, ∠BDA = ∠CDA = 90o

To prove,

ΔABC is isosceles. Proof, In ΔDAB and ΔDAC, ∠BDA = ∠CDA = 90o DA = DA (common) ∠DAB = ∠DAC (from 1) By ASA congruence property, ΔDAB ≅ ΔDAC AB = AC

Hence,

ΔABC is isosceles. !!!