Question from http://www.sgforums.com/forums/2297/topics/359768
In the triangle ABC, BF and CE are perpendicular to AC and AB respectively. D and G are the midpoints of EF and BC respectively. Prove that GD is perpendicular to EF.
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Answer:
dot a perpendicular line from G to BE and label it as H
dot a perpendicular line from G to FC and label it as I
Note that by similar triangles (or by intercept theorem),
Considering triangles HBG and EBC
because BG = ½BC,
EH = ½BE
HG = ½EC
Considering triangles GIC and BFC
because BG = ½BC,
FI = ½FC
GI = ½BF
Thus,
EG² = EH² + HG²
= ¼BE² + ¼EC²
= ¼BC²
GF² = GI² + FI²
= ¼BF² + ¼CF²
= ¼BC²
= EG²
Thus, GF = EG
Since GF = EG, and ED = DF, then GEF is an isosceles triangle where GD is perpendicular to EF
