ABCD is a parallelogram - Leaving Cert Mathematics - Question 6B - 2012

From the properties of cyclic quadrilaterals, we know that the opposite angles sum up to 180 degrees. Thus, we state:

• $|ABC| + |CPA| = 180^{ ext{o}}$.
• $|DPC| + |CBA| = 180^{ ext{o}}$.

This indicates that $|ABP|$ and $|CDP|$ are alternate segments.

We can equate the lengths using the isosceles triangle theorem. Since angles at $P$ are equal:

• $|CD| = |CP|$ due to structure of triangle $CDP$ being isosceles or resulting from chords in equal arcs.

Thus we have shown that:

$|CD| = |CP|$.

This completes the proof.