Circle Geometry =( (1 Viewer)

[U MAD BRO]

I can do part (i) it is just equiangular...
But I can't do the rest
If I try I might get them but I am so tired right now and this is the last question in the paper !

 

[U MAD BRO]

Bump!

 

[barbernator]

ii) as ADM|||CBM,

CP/CM=AE/AM
angle C = angle A.
therefore they are similar 2 sides in the same ratio and their included angle is equal.

might add some more as i get em

iii) angle OFB=90 (midpoint of a chord forms a right angle with the origin)
angle OMP = angle OMQ = angle QEX (simirarry)
therefore they are cyclic quads as opposite angles are supplementary

 

[U MAD BRO]

ii) as ADM|||CBM,

CP/CM=AE/AM
angle C = angle A.
therefore they are similar 2 sides in the same ratio and their included angle is equal.

might add some more as i get em

Or you could say they are equiangular since corresponding angles are subtended by the same arcs.
Part (i) is easy, I can't do the rest

 

[Demento1]

I might scan up my solutions if I can. Typing it all out on LaTeX is time consuming.

 

[barbernator]

Or you could say they are equiangular since corresponding angles are subtended by the same arcs.
Part (i) is easy, I can't do the rest

that is part ii) that I did?

 

[U MAD BRO]

that is part ii) that I did?

Sorry my bad I see it now