No registration required to write.

4 posts
• Page **1** of **1**

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.

MV Rao

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

- martin123456
**Posts:**12**Joined:**Tue Nov 17, 2009 10:15 am**Reputation:****1**

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?

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

- martin123456
**Posts:**12**Joined:**Tue Nov 17, 2009 10:15 am**Reputation:****1**

4 posts
• Page **1** of **1**

Return to Square, Rectangle, Parallelogram, Trapezoid

Users browsing this forum: No registered users and 1 guest