Problem 3

Let

AB

and

AC

be two distinct rays not lying on the same line, and let

\omega

be a circle with center

O

that is tangent to ray

AC

E

and ray

AB

F

. Let

R

be a point on segment

EF

. The line through

O

parallel to

EF

intersects line

AB

P

. Let

N

be the intersection of lines

PR

and

AC

, and let

M

be the intersection of line

AB

and the line through

R

parallel to

AC

. Prove that line

MN

is tangent to

\omega

Step 4 of 5: Extract the parallel ratio

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\frac{AM'}{M'P}=\frac{NE}{EQ}=\frac{NR}{RP} \implies M'R\parallel AC

Detailed analysis

Since $A$ and $N$ correspond under the similarity of $\triangle M'YP$ and $\triangle EOQ$ , the sides through them are proportional: $\dfrac{AM'}{M'P}=\dfrac{NE}{EQ}$ . As $R$ lies on segment $EF$ and $PO\parallel EF$ , the configuration also gives $\dfrac{NE}{EQ}=\dfrac{NR}{RP}$ , and this ratio equality on line $AB$ is exactly the condition (by the converse of the basic proportionality theorem applied in $\triangle APN$ ) for $M'R\parallel AC$ .