Question

AD is a altitude of an isosceles triangle ABC in which ad is equal to AC such that ad bisects BC and AD bisects angle a​

Answer 1

Answer:

what do we need to answer

Step-by-step explanation:

Answer 2

SOLUTION:-

Given:

AD is an altitude of an isosceles ∆ABC in which AB=AC.

To prove:

(1).AD bisect BC

(2).AD bisect ∠A

Proof:

In right ∆ADB & right ∆ADC,

AB= AC [given]

AD = AD [common]

Therefore,

∆ADB ≅ ∆ADC [R.H.S rule]

BD = CD [c.p.c.t]

So,

=) AD bisect BC.

Therefore,

∆ADB ≅ ∆ADC [proved above in (1)]

∠BAD = ∠CAD [c.p.c.t]

So,

AD bisect ∠A.

Hence,

Proved.

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