Question

in the given figure, angle B >angle C , AQ is the bisector of angle BAC and AP|BC . prove that angle QAP = 1/2 (angle B - angle C)

Answer 1

Given :  ∠B >∠C, AQ is the bisector of ∠BAC  and AP⊥BC

To Find :  Prove that ∠QAP = 1/2 (∠B - ∠C)

Solution:

∠A + ∠B + ∠C = 180°

=>  (∠A + ∠B + ∠C)/2 = 90°

ABP is right angle triangle

hence ∠BAP   =  90° - ∠B

∠BAP  + ∠QAP  = ∠BAQ

∠BAQ =  ∠A/2

=> ∠BAP  + ∠QAP  =  ∠A/2

=> ∠QAP  =  ∠A/2 - ∠BAP

=>  ∠QAP  =  ∠A/2 - (90° - ∠B)

 (∠A + ∠B + ∠C)/2 = 90°

=> ∠QAP  =  ∠A/2 - ( (∠A + ∠B + ∠C)/2 - ∠B)

=> ∠QAP  =    ∠A/2 - ∠A/2  - ∠B/2  -  ∠C/2  + ∠B

=> ∠QAP  =    ∠B/2  -  ∠C/2

=> ∠QAP  =    (1/2) (∠B - ∠C)

QED

Hence Proved

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