If triangle ABC is isosceles with AB=AC and C(O,r) is the incircle of the triangle ABC touching BC at L ,prove that L bisects BC

ΔABC is an isosceles triangle. C(O, r) is the incircle of ΔABC touching BC at I.

C(O, r) is the incircle of ΔABC.

∴ O is the point of intersection of angle bisector of ΔABC.

I.e., OB bisects ∠B and OC bisects ∠C.

In ΔABC,

AB = AC (Given)

∴ ∠C = ∠B (Equal sides have equal angles opposite to them)

⇒ ∠OCI = ∠OBI (OB bisects ∠B and OC bisects ∠C)

InΔOBI and ΔOCI,

∠OIB = ∠OIC (Radius is perpendicular to the tangent at point of contact)

∠OBI = ∠OCI (Proved)

OI = OI (Common)

∴ ΔOBI  ΔOCI (AAS concurrence criterion)

⇒ BI = IC (CPCT)

Thus, I bisects the side BC.

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 tough one

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v×lamda(frequency×wavelegth)
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What Mohil posted is right.. but there is an easier way. Since the circle is an incircle.. that means AB, AC and BC are tangents. Let M be the point of intersection at AB and N be at AC. AB=AC.. and AM = AN (tangents from external point are equal) Therefore BM=CN (subtracting the above two equations) But BM=BL and CN=CL (tangents from external point are equal) Since BM=CN.. Hence BL=CL Therefore L bisects BC.
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Answers is
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What is known as incircle of a triangle Pls help
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What us this vivek!! The question was of maths
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Is*
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Given answer is like is face
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