Problem 1

Chords

AB

and

CD

of a circle intersect at

E

inside the circle. Let

M

be an interior point of

EB

. The tangent at

E

to the circle through

D,E,M

meets

BC

and

AC

F,G

. If

t=AM/AB

, find

EF/EG

Step 2 of 3: Second similarity

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\triangle CEG\sim\triangle BMD,\qquad \frac{EG}{CE}=\frac{MD}{BM}

Detailed analysis

The supplementary angle at E gives $\angle CEG=180^\circ-\angle CEF=\angle BMD$ . Also cyclicity gives $\angle ECG=\angle ACD=\angle ABD=\angle MBD$ . Therefore $\triangle CEG\sim\triangle BMD$ , and $\frac{EG}{CE}=\frac{MD}{BM}$ .