Question

In the given figure, ∠B >∠C, AQ is the bisector of ∠BAC and AP⊥BC. Prove that angle QAP = 1/ 2∠ b- ∠ c​

Answer 1

Answer:

Step-by-step explanation:

In the figure AQ is bisector of ∠BAC

∠BAQ = x=∠A​/2.........(1)

Now ∠A+∠B+∠C=180°

∠A=180°−∠B−∠C

x=∠A/2​=90°−∠B/2​−∠C​/2.......(2)

AD is perpendicular to BC,

∠BAP=90°−∠B.................(3)

Now  ∠PAQ=∠BAQ−∠BAP

=x -(90°−∠B)

Putting the values of  x from  (2)

∠PAQ = 90°−∠B/2​−∠C​/2 - (90°−∠B)

=90°−∠B/2​−∠C​/2 - 90°+∠B

∠PAQ =∠B/2​−∠C​/2

∠PAQ=1/2( ∠B -∠C)

Answer 2

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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