Question

IF THE BISECTOR OF INTERIOR ANGLE ,ANGLE Q AND ANGLE R OF A TRIANGLE PQRMEET AT POINT S , THEN PROVED THAT ANGLE QSR=90°+1/2ANGLE P​

Answer 1

Given :   bisectors of angles ∠Q and ∠R of triangle PQR meet at a point S

To find :  prove that  ∠QSR = 90°+(1/2)∠P

Solution:

in Δ PQR

∠Q + ∠R  + ∠P  = 180°   ( sum of angles of triangle)

=>∠Q + ∠R = 180°  - ∠P   Eq1

in Δ SQR

∠SQR + ∠SRQ  + ∠QSR  = 180°   ( sum of angles of triangle)

∠SQR  = ∠Q/2

∠SRQ =  ∠R/2

=> ∠Q/2 + ∠R/2 + ∠QSR  = 180°

=> (∠Q + ∠R )/2 + ∠QSR  = 180°

Substitute ∠Q + ∠R value from Eq1

=> (180°  - ∠P)/2 + ∠QSR  = 180°

=> 90° - ∠P/2 +  ∠QSR  = 180°

=> ∠QSR  = 90° + ∠P/2

QED

Hence proved

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Answer 2

Answer:

so I can make it to the meeting