## anonymous one year ago PLEASE HELP!!! Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $\overline{AD}$ and $\overline{BC}$, and let $Q$ be the intersection of $\overline{AB}$ and $\overline{CD}$. Prove that the angle bisectors of $\angle DPC$ and $\angle AQD$ are perpendicular.

1. anonymous

2. mathmath333

Let $$ABCD$$ be a cyclic quadrilateral. Let $$P$$ be the intersection of $$\overline{AD}$$ and $$\overline{BC}$$, and let $$Q$$ be the intersection of $$\overline{AB}$$ and $$\overline{CD}$$. Prove that the angle bisectors of $$\angle DPC$$ and $$\angle AQD$$ are perpendicular. question

3. ganeshie8

still here?

4. anonymous

yup

5. ganeshie8

you must be knowing that opposite angles in a cyclic quadrilateral add up to 180

6. anonymous

ohhh i see

7. ganeshie8

As a start, use that to show that $$\angle DAB \cong \angle BCQ$$

8. anonymous

oh i see where you're going! thanks so much! I'll try it out myself first and come back if i need more help!

9. ganeshie8

** 1) By triangle exterior angle theorem, $$\angle DAB\cong \angle P + \angle B$$. 2) Since the opposite angles in a cyclic quadrilateral add up to $$180$$ : $$\angle DAB\cong \angle BCQ$$ 3) From above two steps, we get $$\angle BQC \cong 180-(\angle P+\angle B) - \angle B$$

10. ganeshie8

Okay, I'll let you finish it off.. good luck!

11. phi

for what it is worth, here is a complicated approach