1
Question
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.
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
ANSWER:
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⇒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)
⇒MP=MC Corresponding parts of congruent triangles are equal
⇒QR bisects PC.
ANSWER:
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⇒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)
⇒MP=MC Corresponding parts of congruent triangles are equal
⇒QR bisects PC.
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