Prove a parallelogram inside parallelogram

Question

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.

Answer 1

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. $$$$

Answer 2

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

Answer 3

By symmetry of the construction $AP$ and $CQ$ are parallel, and so are $CP$ and $AQ$.