In ΔPQR, PS ⊥ QR and SR > SQ, show that PR > PQ.
PR > PQ
Construction: Draw PT such that SQ = ST
Proof: In ΔPQS and ΔPTS,
PS = PS ...... [common]
SQ = ST ......[By construction]
∠PSQ = ∠PST ...... [each 90°]
... ΔPQS ≅ ΔPTS ...... [SAS criterion]
...∠1 = ∠2 ...... [c.p.c.t.]
Now in ΔPRT, we have the exterior angle ∠2
... ∠2 = ∠3 + ∠TPR
or ∠2 > ∠3
... ∠1 > ∠3 ...... (∠1 = ∠2)
Now in ΔPQR, since
∠1 > ∠3
... PR > PQ ...... [side opposite to greater angle]
Construction: Draw PT such that SQ = ST
Proof: In ΔPQS and ΔPTS,
PS = PS ...... [common]
SQ = ST ......[By construction]
∠PSQ = ∠PST ...... [each 90°]
... ΔPQS ≅ ΔPTS ...... [SAS criterion]
...∠1 = ∠2 ...... [c.p.c.t.]
Now in ΔPRT, we have the exterior angle ∠2
... ∠2 = ∠3 + ∠TPR
or ∠2 > ∠3
... ∠1 > ∠3 ...... (∠1 = ∠2)
Now in ΔPQR, since
∠1 > ∠3
... PR > PQ ...... [side opposite to greater angle]
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