A tough Geometry Question

Question

Two circles (⊙O and ⊙O’, the radius of ⊙O > the radius of ⊙O’) intersect at Point A and Point B. Line CD, which is their common tangent line at the bottom, is drawn tangent to ⊙O and ⊙O’ at Point C and Point D respectively. Extend CB to intersect AD at E, extend DB to intersect AC at F.

PROVE: DA * DE = DB * DF

Answer

Aha, geometry!

Draw a diagram and follow the steps below.

The trick is to show that the triangles DBE and
triangle DAF are similar.

Proof That triangles DBE and triangle DAF are similar.

Let <ABC> donote angle ABC.
We need to show any two corresponding angles of these traingles are equal.

First angle,
<FDA> = <BDE> because D,B,F collinear and D,E,A collinear.

Second angle,
<DAF> = <DAC> because A,F,C collinear
= (pi) – <ACD> – <ADC> angles in triangle ACD
= (pi) – <ACD> – <ADC> …(1)

Now, separately manipulate,<ACD> and <ADC>.

<ACD> = <ACB> + <BCD> splitting <ACD> with line CB
= <ACB> + <CAB>, tangent property <BCD> = <CAB>
= (pi) – <CBA> angles in triangle ABC
= <ABE> because C,B,E are all colinear

Now, similarly

<ADC> = <ADB> + <BDC> splitting <ADC> with line DB
= <ADB> + <BAD>, tangent property <BDC> = <BAD>
= (pi) – <ABD> angles in triangle ABD

Substituiting these two info into (1), we get,

<DAF> = (pi) – <ACD> – <ADC>
= (pi) – <ABE> – [ (pi) – <ABD> ]
= <ABD> – <ABE>
= <BDE> because BE splits <ABD>

Done. In showing the triangle DAF and triangle DBE are similar.

Hence, by ratio for similar triangles,

DA/DF = DB/DE

Or Rewriting we get

DA * DE = DB * DF

Proved.