Question

Constructed externally on the sides AB, AC of ΔABC are equilateral triangle ABX and ACY. If P, Q, R are the midpoints of AX, AY and BC respectively, then ΔPQR is (a) right angle (b) equilateral (c) isosceles (d) None of these

(a) right angle 

(b) equilateral 

(c) isosceles 

(d) None of these

Answer 1

(b) Equilateral.

Let E and F be the mid-points of AB and AC respectively. Then, by mid-point theorem, 

In ΔBAC, 

ER || AC and ER = \(\frac{1}{2}\) AC = AF 

In ΔCAB, FR || AB and FR = \(\frac{1}{2}\)AB = EA 

∴ AERF is a parallelogram.

∴ ΔABX being an equilateral Δ and AERF a parallelogram, 

PE = AE = RF. Similarly for ΔACY and ||gm AERF 

QF = AF = ER 

Also, ∠PER = ∠PEA + ∠AER 

= 60° + ∠AER                                 ...(i) (opp. ∠s of a prallelogram are equal) 

= 60° + ∠AFR = ∠QFR                   ...(ii)

∴ PE = RF, QF = ER and ∠PER = ∠QFR              ...(iii) 

⇒ ΔEPR ≅ ΔQRF 

⇒ PR = QR and ∠EPR = ∠QRF 

Also, ∠PRQ = ∠ERF – ∠ERP – ∠QRF            ...(iv) 

= 180° – ∠AER – ∠ERP – ∠QRF 

= 180° – ∠AER – ∠ERP – ∠EPR                       (From (iv) 

= 180° – ∠AER – (∠ERP + ∠EPR) 

= 180° – ∠AER – (180° – ∠PER) 

= ∠PER – ∠AER = 60°. 

Also, PR = QR ⇒ ∠RPQ = ∠RQP = \(\frac{1}{2}\) (180° – ∠PRQ) 

= \(\frac{1}{2}\) (180° – 60°) = 60°. 

⇒ ΔPQR is equilateral.