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 1 of 3: First similarity

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\triangle CEF\sim\triangle AMD,\qquad \frac{EF}{CE}=\frac{MD}{AM}

Detailed analysis

Because ABCD is cyclic, $\angle ECF=\angle DCB=\angle DAB=\angle MAD$ . The tangent-chord theorem for the circle DEM gives $\angle CEF=\angle EMD=\angle AMD$ . Hence $\triangle CEF\sim\triangle AMD$ , so $\frac{EF}{CE}=\frac{MD}{AM}$ .