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I have drawn a figure,

In parallelogram $ABCD$, $AP$ is the bisector of angle $A$, $CQ$ is the bisector of angle $C$

Can I prove $APCQ$ is a parallelogram? or it isn't?

I first joined $AC$ and now if somehow I can show $AC$ and $PQ$ bisect each other then I can prove $APCQ$ is a parallelogram.

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I would do it by congruent triangles. You can prove $ADB \cong CBD, ADP \cong CBQ$, and so on to get there.

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  • $\begingroup$ I finally proved DP = BQ , now how do I prove AC bisects PQ? $\endgroup$ – Ashesh Feb 1 '16 at 15:15
  • $\begingroup$ @AsheshKumar You dont need to draw AC. congruency of DPC and BQA will prove the other pair of alternate interior angles equal, so the other pair of lines will be proved parallel too $\endgroup$ – Max Payne Feb 1 '16 at 15:19
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It is a parallelogram. Use the condition for a quadrilateral to be parallelogram, ie sides are parallel. To check this, you just need to check whether alternate interior angles are equal.

Now, DP = QB as APD and CQB are congruent.

PDC will be congruent to QBA (by SAS)

so angle DPC = angle BQA or angle CPQ = angle PQA

So remaining sides are also proved parallel. $$$$

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  • $\begingroup$ I am still a novice , from what you told I proved angle APQ = angle CQP . what's next? $\endgroup$ – Ashesh Feb 1 '16 at 14:52
  • $\begingroup$ Use asa congruency by drawing diagonal $\endgroup$ – Max Payne Feb 1 '16 at 15:01
  • $\begingroup$ How you can prove AQ=PC? $\endgroup$ – Ashesh Feb 1 '16 at 15:20
  • $\begingroup$ You had APD and CQB congruent. Therefore by CPCT. (corresponding prats of congruent triangles.), BQ = PD Now Again, ABQ and CPD are congruent (using SAS, since AB = CD and BQ = PD, and angle ABQ = angle CPD (alt interior angle) ). Therefore AQ = PD $\endgroup$ – Max Payne Feb 1 '16 at 15:30
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By symmetry of the construction $AP$ and $CQ$ are parallel, and so are $CP$ and $AQ$.

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