Question

QI and RI bisect exterior angle RQT and QRS . Prove that angle QIR = 90° - 1 / 2 angle P​

Answer 1

45 I thought this is correct answer

Answer 2

Given :  ΔPQR and  PQ is Extended to T  and  PR is extended to S

QI and RI bisect exterior angle RQT and QRS

To Find :  Prove that  ∠QIR = 90° - (1/2)P

Solution:

Sum of angle of a triangle = 180°

Exterior angle of a Triangle = Sum of opposite two internal angles of triangle

Angle bisector divided angle in 2 Equal measures

in Δ QIR

=> ∠QRI + ∠RQI  + ∠QIR = 180°

=> (1/2) ∠QRS + (1/2)∠RQT  + ∠QIR = 180°

=> (1/2) (   ∠QRS + ∠RQT) + ∠QIR = 180°  

∠ QRS = ∠Q + ∠P

∠RQT = ∠R + ∠P

=> (1/2) (   ∠Q + ∠P + ∠R + ∠P) + ∠QIR = 180°  

∠Q + ∠P + ∠R = 180°

=> (1/2) (   180° + ∠P) + ∠QIR = 180°  

=> 90° + (1/2)∠P + ∠QIR = 180°  

=>  (1/2)∠P + ∠QIR =90°  

=>  ∠QIR = 90°    -  (1/2)∠P

QED

Hence proved

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