Given: P is the centre of the circle.The Problem:

CE is the diameter.

∠ 1= 35°

EAB = 100°

I found the measure of every angle, with the exception of ∠5.Partial Solution:

Here's how I found the other angles.

The inscribed angle is half the intercepted arc. So ∠ECB = 1/2 (EAB)

∠EBC = 1/2 (100°) = 50°

EC, as a diameter, cuts the circle in two, making each half 180°.

So EAB + BC = 180°

∠3 is 1/2 (BC) as the inscribed angle is half the intercepted arc.

So ∠3 = 1/2(80°) = 40°

Consider triangle EBC, with interior angles adding up to 180°.

180° -40° (∠3) -50° (∠ECB) -35° (∠1) =∠2

∠2 = 55°

Consider semi-circle EDC.

EP bisects the semicircle in two 90° halves. So DC = 90°

Back to the whole circle, inscribed angle ∠4 is half intercepted arc DC, so ∠4 is 45°.

From here I'm stuck,as I can't find ∠5.