Question

In the given figure, PR > PQ and PS bisects angle QRP. Prove that angle PSR > angle PSQ.​

Answer 1

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{Solution:-}$

⏩ In ∆ PQR, we have

PR > PQ (Given)

angle PQR > angle PRQ

(Since, angle opposite to the larger side is greater)

Therefore,

angle PQR + angle 1 > angle PRQ + angle 1

(Adding angle 1 on both sides)

angle PQR + angle 1 > angle PRQ + angle 2 ....→ (i)

(Since, PS is the bisector of angle P , therefore, angle 1 = angle 2)

Now,

In ∆PQS and ∆PSR, we have,

angle PRQ + angle 1 + angle PSQ = 180° (Angle sum property of a triangle)

and

angle PRQ + angle 2 + angle PSR = 180°

angle PQR + angle 1 = 180° - angle PSQ ...→ (ii)

and

angle PRQ + angle 2 = 180° - angle PSR ...→ (iii)

→ Substitute (ii) and (iii) in eq. (i)

since,. 180° - angle PSQ > 180° - angle PSR

=> - angle PSQ > - angle PSR

=> angle PSQ < angle PSR

or

angle PSR > angle PSQ

_____❗ Hence Proved ❗_____

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Hope it helps...:-)

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