Question

PA and PB are the tangents to a circle which circumscribes an epuilateral triandleABQ . if angle PAB 60 degree .prove that triangle ABP is an equilateral triangle according to class 10 cbse

Answer 1

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Given,ΔAQB is equilateral. PA and PB are tangents drawn from P.To prove: PQ perpendicular bisector of AB.Proof:We have in ΔPAQ and ΔPBQAQ = QB [∵sides of equilateral triangle are equal]PQ = PQ [∵ common side]PA = PB [∵ Tangents drawn from a point to a circle are equal]ΔPAQ ≅ ΔPBQ [∵SSS SSS congruence condition]∠PQA=∠PQB [∵CPCT]In ΔACQ and ΔBCQAQ = QB [∵sides of equilateral triangle are equal]∠PQA=∠PQB CQ = CQ [∵common side]ΔACQ ≅ ΔBCQ AC = CB [∵CPCT]Thus, QC median of equilateral ΔAQB.QC⊥AB [∵Median of an equilateral triangle is its perpendicular bisector ]So QP is perpendicular bisector of ABProved.

Answer 2

Equilateral Triangle Possess Equal Sides

Step-by-step explanation:

Given,

ΔAQB is equilateral. PA and PB are tangents drawn from P.

To prove : PQ perpendicular bisector of AB.

Proof : We have in ΔPAQ and ΔPBQ

• AQ = QB             [∵ sides of equilateral triangle are equal]

• PQ = PQ              [∵ common side]

• PA = PB              [∵ Tangents drawn from a point to a circle are equal]

ΔPAQ ≅ ΔPBQ [∵SSS congruence condition]

• ∠PQA=∠PQB    [∵CPCT]

In ΔACQ and ΔBCQ

• AQ = QB [∵ sides of equilateral triangle are equal]

• ∠PQA=∠PQB  

• CQ = CQ [∵ common side]

ΔACQ ≅ ΔBCQ  

AC = CB [∵ CPCT]

Thus, QC median of equilateral ΔAQB.

QC⊥AB [∵ Median of an equilateral triangle is its perpendicular bisector ]

So, QP is perpendicular bisector of AB.