Question

AD is an altitude of an isosceles triangle ABC in which AB=AC show that. (i)AD bisects BC (ii) AD bisect LA​

Answer 1

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Given -

➩ AD is an altitude

➩ AB = AC

To find -

➩ AD bisects BC

➩ AD bisects $\angle$ A

Solution -

In Triangle ABD & Triangle ACD

➩ AB = AC [given]

➩ $\angle$ B = $\angle$ C [angles opposite to the equal sides of a Triangle are equal]

➩ AD = AD [common]

$\therefore$ Triangle ABD is congurent to Triangle ACD [ASA rule]

So,

AD Bisects BC [CPCT]

AD Bisects $\angle$ A [CPCT]

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Answer 2

Given -

➩ AD is an altitude

➩ AB = AC

To find -

➩ AD bisects BC

➩ AD bisects ∠ A

Solution -

In Triangle ABD & Triangle ACD

➩ AB = AC [given]

➩ ∠ B = ∠ C [angles opposite to the equal sides of a Triangle are equal]

➩ AD = AD [common]

∴ Triangle ABD is congurent to Triangle ACD [ASA rule]

So,

AD Bisects BC [CPCT]

AD Bisects BC [CPCT]AD Bisects ∠ A [CPCT]

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