Let AA' and CC' be medians of ΔABC intersecting in point G. Let m be any line through G intersecting sides AB and AC.
Let P, Q, and R be the feet of perpendiculars to m from
A, B, and C respectively.

Prove that |AP| = |BQ| + |CR|.

Let S be the foot of the perpendicular from A’ to m.

Triangles APG and A’SG are similar (AAA)

Since AG = 2 A’G, (centroid property), it follows that         AP = 2 A’S         (1)

Let the line through S parallel to BC cut BQ and CR at U and V respectively, thus forming congruent parallelograms A’BUS and CA’SV.

Triangles SQU and SRV are congruent (SAA), therefore     QU = RV            (2)

(1) now gives:         AP = A’S + A’S
                                    = BU + CV
                                    = (BQ + QU) + (CR - RV)
                                    = BQ + CR                    using (2)

Posted by Harry on 2010-08-15 17:15:15