short and simple circle geo question. (1 Viewer)

[conics2008]

The red drawings are what I drew.

I can say with confidence that CD is diameter

PCKA = Cyclic Quad

< b = 90

but still cant to seem to get in KBL and DKL

help please.

 

[Iruka]

Is L the point of intersection of KP and BD?

If so, then you prove that triangle BKL is similar to triangle KPA, using the fact that PCKA and CBDA are cyclic quadrilaterals.

Angle BKE = angle PKA (vert opp) then you just have to show that

angle KBE = angle APK

so the other angles are equal also.

 

[conics2008]

hey where is point E ??????

 

[peckerhead]

let angle CBA =x
then angle PCA = x (alternate segment, a tangent at C?)
and because CKAP is cyclic
then angle PKA = x

that's all you need: CB and PK are both inclined at same angle to AB (ie x) and are thence parallel.

result follows.