In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

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Solution

$Since ∆ ABC is an equilateral ∆, then ∠ ABC = ∠ BCA = ∠ CAB = 60 ° Since PQ ∥ CA and PC is a transversal, then ∠ BPQ = ∠ BCA = 60 ° (Corresponding angles) Since PQ ∥ CA and QA is a transversal, then ∠ BQP = ∠ BAC = 60 ° (Corresponding angles) Further, ∠ B = 60 ° In ∆ BPQ, ∠ B = ∠ BPQ = ∠ BQP = 60 ° ∴ △ BPQ is an equilateral triangle . i . e ., BP = PQ = BQ Now, BP = CR \Rightarrow PQ = CR . . . . (1) Considering △ MPQ and △ MCR, we get : ∠ PQM = ∠ MRC (Alternate interior angles) ∠ PMQ = ∠ CMR (Vertically opposite angles) PQ = CR [using (1)] ∆ MPQ ≅ ∆ MCR (AAS criterion) \Rightarrow MP = MC [Corresponding parts of congruent triangles are equal] \Rightarrow QR bisects PC .$

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