Prove that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal.

(i) ∠TRQ = ∠SQR?

(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.

(iii) Is ΔPQR an equilateral triangle?

In ΔQTR and ΔRSQ

∠QTR = ∠RSQ = 90° (Given)

∠TQR = ∠SRQ (Base angle of an isosceles triangle)

∠QRT = ∠RQS (Remaining third angles)

QR = QR (Common)

ΔQTR = ΔRSQ (By ASA rule)

QS = RT (Congruent parts of congruent triangles)

Hence proved.

(i) ∠TRQ = ∠SQR (Congruent parts of congruent triangles)

(ii) In ΔQTR,

∠TRQ = 30° (Given)

∠QTR + ∠TQR + ∠QRT = 180° (Angle sum property)

⇒ 90° + ∠TQR + 30° = 180°

⇒ 120° + ∠TQR = 180°

⇒ ∠TQR = 180° – 120° = 60°

⇒ ∠TQR = ∠SRQ = 60°

Each base angle = 60°

(iii) In ΔPQR,

∠P + ∠Q + ∠R = 180° (Angle sum property)

⇒ ∠P + 60° + 60° = 180° (From ii)

⇒ ∠P + 120° = 180°

⇒ ∠P = 180° – 120° = 60°

Hence, ΔPQR is an equilateral triangle.