And because AP and BP are bisectors, ∠ PAB+∠ PBA =1/2 (∠ QAB+∠ ABC) =1/2 *180 = 90 = "∞.

That is, △APB is a right triangle.

2. Because AP and BP are bisectors, ∠ DAP = ∠ PAB, ∠ AP, BP = ∠ CBP.

Because DC//AB, ∠ DPA = ∠ PAB = ∠ DAP = ∠ AD = DP.

Similarly: BC=PC because BC=AD and DP=PC.

3. Because AB is the diameter, ∠ AEB = 90, that is ∠ AEF = 90, ∠DPA=∠PAB.

So △AEF is similar to△ △APB.

Because ∠ APB = 90, P is on the circle O, even PO.

OP = AD = 5 => From Pythagorean theorem, AB =10: Pb = √ (102-82) = 6.

tanAFE=tanABP=AP/BP=8/6=4/3