In Fig. 10.23, PQRS is a square and SRT is an equilateral triangle. Prove that

(i) PT = QT (ii) ∠TQR =15°

Given,

PQRS is a square and SRT is a equilateral triangle

To prove: (i) PT = QT

(ii) ∠TQR =15°

Proof: PQ = QR = RS = SP (Given) (i)

And, ∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90^o

And also,

SRT is a equilateral triangle

SR = RT = TS (ii)

And, ∠TSR = ∠SRT = ∠RTS = 60^o

From (i) and (ii)

PQ = QR = SP = SR= RT = TS (iii)

∠TSP = ∠TSR + ∠RSP

= 60^o + 90^o = 150^o

∠TRQ = ∠TRS + ∠SRQ

= 60^o + 90^o = 150^o

Therefore, ∠TSR = ∠TRQ = 150^o (iv)

Now, in and , we have

TS = TR (From iii)

∠TSP = ∠TRQ (From iv)

SP = RQ (From iii)

Therefore, By SAS theorem,

PT = QT (BY c.p.c.t)

In

QR = TR (From iii)

Hence, is an isosceles triangle.

Therefore, ∠QTR = ∠TQR (Angles opposite to equal sides)

Now,

Sum of angles in a triangle is 180^o

∠QTR + ∠TQR + ∠TRQ = 180^O

2∠TQR + 150^O = 180^O (From iv)

2∠TQR = 30^O

∠TQR = 15^O

Hence, proved