Question
Sat March 14, 2015 By: Vatsal Choudhary

In the previous question asked by me PQRS is a parallelogram. QA bisects angle pqr and meet PS at A. A straight line through R and parallel to QA meets PQ produced at B and PS produced at C. Prove that traingle PBC is an isosceles triangle. The solution provided by topperlearning experts is totally wrong. The question is asking something and the answer provided is telling different thing. Infact the solution provided by your experts is wrong. I have my exam today and I posted that question yesterday. I thought that I will get the correct answer for it. But I couldn't get that. Sorry to say, but my trust over topperlearning experts has been lifted.

Expert Reply
Vimala Ramamurthy
Sun March 15, 2015
Consider the following figure.
I n space a space p a r a l l e log r a m comma space a d j a c e n t space a n g l e s space a r e space s u p p l e m e n t a r y. T h u s comma angle Q P S plus angle P Q R equals 180 degree G i v e n space t h a t space Q A space i s space b i s e c t i n g space angle P Q R. rightwards double arrow angle P Q A equals angle A Q R equals fraction numerator space angle P Q R over denominator 2 end fraction A l s o space f r o m space t h e space s t a t e m e n t space t h a t space B R space i s space p a r a l l e l space t o space Q A.  T h u s comma B Q parallel to R S space a n d space Q R space i s space t h e space t r a n s v e r s a l.  T h e r e f o r e comma space angle A Q R equals fraction numerator space angle P Q R over denominator 2 end fraction comma space rightwards double arrow angle A Q R equals angle Q R B equals fraction numerator space angle P Q R over denominator 2 end fraction.... left parenthesis 1 right parenthesis  B Q parallel to R S space a n d space Q R space i s space t h e space t r a n s v e r s a l. rightwards double arrow angle A P B equals angle R Q B... left parenthesis 2 right parenthesis C o n s i d e r space t h e space t r i a n g l e comma space B Q R B y space a n g l e space s u m space p r o p e r t y angle R Q B plus angle Q R B plus angle Q B R equals 180 degree rightwards double arrow angle A P B plus fraction numerator space angle P Q R over denominator 2 end fraction plus angle Q B R equals 180 degree space space left square bracket f r o m space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis right square bracket rightwards double arrow angle Q B R equals 180 degree minus angle A P B minus fraction numerator space angle P Q R over denominator 2 end fraction B Q parallel to R S space a n d space h e n c e comma space angle Q B R equals angle S R C... left parenthesis 3 right parenthesis P Q parallel to R S space a n d space h e n c e comma space angle A P B equals angle R S C... left parenthesis 4 right parenthesis   C o n s i d e r space t h e space t r i a n g l e space S R C comma B y space a n g l e space s u m space p r o p e r t y comma space w e space h a v e angle R S C plus angle S R C plus angle P C B equals 180 degree rightwards double arrow angle A P B plus 180 degree minus angle A P B minus fraction numerator space angle P Q R over denominator 2 end fraction plus angle P C B equals 180 degree space space space space space space left square bracket F r o m space left parenthesis 3 right parenthesis space a n d space left parenthesis 4 right parenthesis right square bracket rightwards double arrow angle P C B equals angle A P B plus fraction numerator space angle P Q R over denominator 2 end fraction  C o n s i d e r space t h e space t r i a n g l e space P B C angle Q B R equals 180 degree minus angle A P B minus fraction numerator space angle P Q R over denominator 2 end fraction comma space angle A P B space a n d space angle A P B plus fraction numerator space angle P Q R over denominator 2 end fraction E i t h e r space d a t a space i s space i n c o m p l e t e space o r space p l e a s e space p r o v i d e space a space p r o p e r space q u e s t i o n space w i t h space d i a g r a m
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