Q. 174.0( 4 Votes )

# An equilateral triangle ABC is inscribed in a circle. P is any point on the minor arc BC. Prove that PA = PB + PC.

Answer :

Equilateral ΔABC inscribed in a circle C(O, r) with point P on minor arc BC.

To Prove: PA = PB + PC

Construction: Produce BP to Q such that PQ = PC. Join C to Q.

Proof: ABPC is a cyclic quadrilateral .

But ∠A = 60° [angle of a Equilateral triangle]

.

.

= 180° - 120°

= 60° .............. (ii)

Now in ΔPQC,

∠3 + ∠4 + ∠Q = 180°

.

But ∠3 = ∠Q

.

Hence, ΔPQC is equilateral

∴ ∠2 = ∠3 = 60°

∴ ∠2 + ∠BCP = ∠3 + ∠BCP

⇒ ∠ACP = ∠BCQ

Now in ΔACP and ΔBCQ, ∠1 = ∠5 [Angles in the same segment of the circle]

∠ACP = ∠BCQ [proved above]

and PC = QC [by construction]

.

.

⇒ AP = BP + PC [PC = PQ].

To Prove: PA = PB + PC

Construction: Produce BP to Q such that PQ = PC. Join C to Q.

Proof: ABPC is a cyclic quadrilateral .

^{.}. ∠A + ∠BPC = 180° .............. (i)But ∠A = 60° [angle of a Equilateral triangle]

.

^{.}. ∠BPC = 180° - 60° = 120°.

^{.}. ∠CPQ = 180° - ∠BPC= 180° - 120°

= 60° .............. (ii)

Now in ΔPQC,

∠3 + ∠4 + ∠Q = 180°

.

^{.}. ∠3 + ∠Q = 120°But ∠3 = ∠Q

.

^{.}. ∠3 = ∠Q = 60°Hence, ΔPQC is equilateral

∴ ∠2 = ∠3 = 60°

∴ ∠2 + ∠BCP = ∠3 + ∠BCP

⇒ ∠ACP = ∠BCQ

Now in ΔACP and ΔBCQ, ∠1 = ∠5 [Angles in the same segment of the circle]

∠ACP = ∠BCQ [proved above]

and PC = QC [by construction]

.

^{.}. ΔACP @ DBCQ.

^{.}. AP = BQ = BP + PQ⇒ AP = BP + PC [PC = PQ].

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