Circle geometry question (1 Viewer)

[porcupinetree]

For part iii, is it enough to do the following proof?

I spent ages trying to look for a way to prove that opposite angles are supplementary (this is usually what questions like this want), but couldn't figure it out.

 

[Drongoski]

View attachment 31950
For part iii, is it enough to do the following proof?
View attachment 31951
I spent ages trying to look for a way to prove that opposite angles are supplementary (this is usually what questions like this want), but couldn't figure it out.

iii)

Having established ii)

- angle BPD = angle BQD = angle AQE = @ say.

- .: angle PBD = 90 - @

But angle PAC = angle PBD = 90 - @ (both angles stand on minor arc PC)

In triangle AQE, angle AEQ = 180 - @ -(90 - @) = 90

.: angle BDA = angle BEA = angle AEQ

Both angles stand on same line segment AB and are on the same side of AB

.: ABDE is cyclic

iv) we have already shown angle BEA = 90


Edit

Sorry I did not see your proof; it is OK.

Also: in my proof, I should have pointed out that angle DBQ = angle DBE = angle DAE

and both angles stand on DE and are on the same side of DE.

.: ABDE is cyclic.

 

[porcupinetree]

iii)

Having established ii)

- angle BPD = angle BQD = angle AQE = @ say.

- .: angle PBD = 90 - @

But angle PAC = angle PBD = 90 - @ (both angles stand on minor arc PC)

In triangle AQE, angle AEQ = 180 - @ -(90 - @) = 90

.: angle BDA = angle BEA = angle AEQ

Both angles stand on same line segment AB and are on the same side of AB

.: ABDE is cyclic

iv) we have already shown angle BEA = 90


Edit

Sorry I did not see your proof; it is OK.

Also: in my proof, I should have pointed out that angle DBQ = angle DBE = angle DAE

and both angles stand on DE and are on the same side of DE.

.: ABDE is cyclic.

Thank you heaps