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Question
In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.
Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).
Solution
ΔABC is an equilateral triangle,
So, each of its angles equals 60°.
QP is parallel to AC,
⇒ ∠PQB = ∠RAQ = 60°
ln ΔQBP,
∠PQB = ∠BQP = 60°
So, ∠PBQ + ∠BQP + ∠BPQ = 180° ....(angle sum property)
⇒ 60°+ 60° + ∠BPQ = 180°
⇒ ∠BPQ = 60°
So, ΔBPQ is an equilateral triangle.
⇒ QP = BP
⇒ QP = CR ....(i)
Now, ∠QPM + ∠BPQ = 180° ...(linear pair)
⇒ ∠QPM+ 60°= 180°
⇒ ∠QPM = 120°
Also, ∠RCM+ ∠ACB = 180° ...(linear pair)
⇒ ∠RCM+ 60° = 180°
⇒ ∠RCM = 120°
ln ΔRCM and ΔQMP,
∠RCM = ∠QPM ....(each is 120°)
∠RMC = ∠QMP ...(vertically opposite angles)
QP= CR ....(from(i))
⇒ ΔRCM ≅ ΔQMP ....(AAS congruence criterion)
So, CM = PM
⇒ QR bisects PC.
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