Proving a figure in trapezium as parellologram

Proving a figure in trapezium as parellologram

Postby MV RAO » Wed Dec 09, 2009 5:10 am

ABCD is a trapezium with AB parallel to CD. Angle bisector of A meets DC at P and angle bisector of C meets AB at Q. Prove that APCQ is a parellogram.
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Re: Proving a figure in trapezium as parellologram

Postby martin123456 » Thu Dec 10, 2009 1:45 pm

MV RAO wrote:ABCD is a trapezium with AB parallel to CD. Angle bisector of A meets DC at P and angle bisector of C meets AB at Q. Prove that APCQ is a parellogram.

let [tex]\angle{A}=2\alpha[/tex]. AB||CD =>[tex]\angle{DPA}=\alpha[/tex]. Analog, let [tex]\angle{BCD}=2\beta[/tex]=>[tex]\angle{CQB}=\beta[/tex]. we have to prove that [tex]\alpha+\beta=\pi[/tex].
i think this is not true

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Re: Proving a figure in trapezium as parellologram

Postby MV RAO » Thu Dec 10, 2009 11:25 pm

martin123456 wrote:
MV RAO wrote:ABCD is a trapezium with AB parallel to CD. Angle bisector of A meets DC at P and angle bisector of C meets AB at Q. Prove that APCQ is a parallelogram.

let [tex]\angle{A}=2\alpha[/tex]. AB||CD =>[tex]\angle{DPA}=\alpha[/tex]. Analog, let [tex]\angle{BCD}=2\beta[/tex]=>[tex]\angle{CQB}=\beta[/tex]. we have to prove that [tex]\alpha+\beta=\pi[/tex].
i think this is not true


This is a text book problem for CBSE 8th standard in India. So I guess the problem is correct. To prove tha APCQ is a parallelogram, I believe if we can prove [tex]\alpha = \beta[/tex], it will be sufficient. How to prove it?

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Re: Proving a figure in trapezium as parellologram

Postby martin123456 » Fri Dec 11, 2009 8:28 am

well, it is sufficient but not true, try drawing a trapezoid that fulfills the conditions given but APCQ not a parallelogram

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