In □PQRS, = {T}, PS = QR,. Prove that ΔTPS is similar to ΔQTR.

Given: In PQRS, PR ∩ QS = {T}, PS = QR and PQ||RS.

To prove: ∆TPS and ∆QTR are similar.

Construction: Through R, draw a line parallel to PS to intersect PQ at M.

Proof:

In PMRS, PM||RS (P-M-Q) and PS||MR (construction).

Thus, PMRS is a parallelogram.

PS = MR

Further, PS = QR (given)

In ∆RMQ, ∠RMQ ≅ ∠RQM …(i)

Further, the corresponding angles formed by transversal PQ to PS||MR are congruent.

∠RMQ ≅ ∠SPM (corresponding angles)

∠RQM ≅ ∠SPM (by (i))

∠RQP ≅ ∠SPQ (P-M-Q)

In ∆SPQ and ∆RQP,

∠SPQ ≅ ∠RQP

PQ ≅ QP (same line segment)

SP ≅ RQ (given)

The correspondence ∆SPQ↔∆RQP is a congruence by SAS theorem of congruence.

∠PSQ ≅ ∠QRP

∠PST ≅ ∠QRT (S-T-Q and R-T-P)

In ∆TPS and ∆QTR

∠STP ≅ ∠RTQ (Vertically opposite angles)

∠PST ≅ ∠QRT

∆TPS and ∆QTR are similar by AA corollary.

Thus, by AA corollary the correspondence ∆TSP↔∆TRQ is a similarity.