Stuck (1 Viewer)

[InteGrand]

a) ∠AEB = ∠ABE (base angles of isosceles triangle ABE)
= ∠ACE (angles on same arc AE). Q.E.D.
b) ∠BQC = ∠AQE (vertically opposite angles)
= ∠QAB + ∠ABQ (external angle of triangle AQB)
= y + x (due to our definitions of x and y, and ∠ABQ = ∠ABE = x)
= x + y (symmetric property of equality). Q.E.D.
c) ∠ACB = ∠AEB (angles on same arc AB) = x.

∴ ∠ABC = 180° – (x + y) (angle sum of triangle ABC)
∴ ∠PDC = x + y (opposite angles of cyclic quadrilateral ADCB are supplementary)
⇒ ∠PDC = ∠BQC
⇒ PQCD is a cylic quadrilateral (interior angle equals exterior opposite angle). Q.E.D.

 

[Speed6]

Nah its okay guys, I like the team effort here. I will post a question or two up in a second.
I would like to thank everyone thats helping me atm especially Ekman.
I've got my extension math exam on Tuesday and maths test on Wednesday.

 

[Speed6]

 

[Ekman]

a) Angle in alternate segment
b) Since XBA = AYZ (exterior angle in cyclic quad), thus CXA=AYZ. Since these two angles are equal, lines CD and YZ must be parallel due to alternate angles in parallel lines hold true.
c)A-Angle X is shared, A-XBA=XYZ, A-XZY=XAB. Therefore due to AAA, the triangles are similar.
d)AX/6 = 4/(AX+5) (Ratio of sides in similar triangles proven in part (c)). Solve and you get AX = 3