I would use similar triangles
First ACP ~ AFQ (use alt angle of parallel lines, and one common angle, triangles with three same angles are sim)
Then, AF=FC given midpoint.
which then, using proportion, aq=qp
In a triangle ABC,E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects EF at Q.Prove AQ =PB.
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