Question

In the given figure ABC is an isosceles triangle in which AB equal a Also D is a point such that BD equal CD. prove that ad bisects angle A and angle d​

Answer 1

Consider:-

Let the intersection point of AD & BC be X

So that not to use D in case of upper triangle and A in case of lower triangle.

To Prove:-

• AD bisects angle A & angle D

that is angle BAX=angle CAX & angle BDX=angle CDX

Solution:-

To prove

angle BAX=angle CAX

Consider ∆BAX & ∆CAX

AX=AX(common)

AB=AC(given isosceles)

Angle ABX=angle ACX (angles opposite to equal sides)

Therefore by S.A.S congruence

∆BAX≈∆CAX

Thus

angle BAX=angle CAX (C.P.C.T)

AD bisects angle A

Now ,Consider ∆BDX & ∆CDX

DX=DX (common)

BD=CD (given isosceles)

angle DBX=angle BCX (angles opposite to equal sides)

Therefore by S.A.S congruence

∆BDX≈∆CDX

Thus

angle BDX=angle CDX

AD bisects angle D

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