Internal Intersection Of Two Circles

[Soumyadeep mandal]

S_1 and S_2 are two circles touching internally at O, with S_1being the inner circle. A straight line cuts S_1 at A, D and S_2 at B, C. prove that :

AB:CD = (OA.OB):(OC.OD)

I need a rigorous proof for this please... thank you.

[Nitin Prasad]

The following figure illustrates the given situation, and we need to show that  $$\frac{AB}{CD}=\frac{OA.OB}{OC.OD}$$

Observe that $$\frac{AB}{CD}=\frac{ar(OAB)}{ar(OCD)}=\frac{\frac{1}{2}OA\cdot OB\sin(\angle AOB)}{\frac{1}{2}OC\cdot OD\sin(\angle COD)}$$

Hence if $$\angle AOB=\angle COD$$ then we are done.
Now observe that $$\angle EOC+\angle COD=\angle EOD=\angle OAD=\angle OBA+\angle AOB=\angle EOC+\angle AOB$$
Hence we have $$\angle AOB=\angle COD$$

Here you should consider directed angles to avoid configuration issues

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