Question

in the given figure PQ=PR and TQ=TR then show that (1) PS bisects angle QTR. (2) PS is perpendicular to QR​

Answer 1

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Answer 2

Given :- the given figure PQ = PR and TQ = TR.

To Prove :-

• (i) PS bisects angle QTR .
• (ii) PS is perpendicular to QR .

Solution :-

(i)

In ∆PQT and ∆PRT we have,

→ PQ = PR { given }

→ TQ = TR { given }

→ PT = PT { common }

So,

→ ΔPQT ≅ΔPRT {By SSS congruence rule }

then,

→ ∠QPT = ∠RPT { By CPCT }

Or,

→ ∠QPS = ∠RPS ---------- Equation (1)

Therefore, we can conclude that, PS is angle bisector of ∠QPR .

Now, Since PQR is an isosceles ∆ as PQ = PR . Then, angle bisector from vertex also bisect the base of the isosceles ∆ .

hence,

→ QS = RS ------------- Equation (2)

Now, In ∆TQS and ∆TRS we have,

→ TQ = TR { given }

→ QS = RS { from Equation (2) }

→ TS = TS { common }

So,

→ ΔTQS ≅ΔTRS {By SSS congruence rule }

then,

→ ∠QTS = ∠RTS { By CPCT }

therefore,

→ TS bisects angle QTR .

Or,

→ PS bisects angle QTR (Proved) .

(ii)

In ∆QPS and ∆RPS we have,

→ PQ = PR { given }

→ ∠QPS = ∠RPS { from Equation (1) }

→ PS = PS { common }

So,

→ ΔTQS ≅ΔTRS {By SAS congruence rule }

then,

→ ∠QSP = ∠RSP { By CPCT } -------- Equation (3)

now, since QR is a straight line,

→ ∠QSP + ∠RSP = 180° { Linear pair }

using Equation (3),

→ ∠QSP + ∠QSP = 180°

→ 2•∠QSP = 180°

→ ∠QSP = 90°

therefore, we can conclude that,

→ PS is perpendicular to QR (Proved) .

Remember :-

• In Equation (1) we proved that, PS is angle bisector of isosceles ∆QPR . Then, PS is also perpendicular to QR by isosceles triangle theorem .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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