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 3 of 3: Divide the ratios and substitute t

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\frac{EF}{EG}=\frac{BM}{AM}=\frac{(1-t)AB}{tAB}=\frac{1-t}{t}

Detailed analysis

Dividing the two similarity relations cancels CE and MD, giving $\frac{EF}{EG}=\frac{BM}{AM}$ . Since M lies on AB, $AM=tAB$ and $BM=(1-t)AB$ . Therefore $\frac{EF}{EG}=\frac{BM}{AM}=\frac{(1-t)AB}{tAB}=\frac{1-t}{t}$ .