Question 1024786: Given:ABCD is a parallelogram, DE is perpendicular to AC and BF is also perpendicular to AC. Prove AE is congruent to FC
Answer by ikleyn(52858)   (Show Source):
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Given:ABCD is a parallelogram, DE is perpendicular to AC and BF is also perpendicular to AC. Prove AE is congruent to FC
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Use the fact that the diagonal of the parallelogram divides it in two congruent triangle.
Hence, the triangles ACB and ACD have the same area.
They also have the common base AC.
Therefore, the altitudes of these triangles, AE and FC, are congruent.
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