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Question
Prove that a triangle ABC is isosceles, if: altitude AD bisects angles BAC.
Solution
In ΔABC, let the altitude AD bisects ∠BAC.
Then we have to prove that the ΔABC is isosceles.
In triangles ADB and ADC,
∠BAD = ∠CAD ...(AD is bisector of ∠BAC)
AD = AD ...(common)
∠ADB = ∠ADC ....(Each equal to 90°)
⇒ ΔADB ≅ ΔADC ...(by ASA congruence criterion)
⇒ AB = AC ...(cpct)
Hence, ΔABC is an isosceles.
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