O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/361016

In the trapezium ABCD, AD is parallel to BC, BC = 3AD and the diagonals AC and BD meet at E. The line through A is drawn parallel to DB to meet the extended line of CB at F. Prove that

(a) FB = AD
(b) EC = 3AE

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Answer:

(a) Since AF is parallel to BD, and AD is parallel to BC (because F is an extension of BC, hence making both FB and FC also parallel to AD ), ADBF must be a parallelogram, with FB = AD.

(Extra notes: The opposite sides of a parallelogram are always of equal lengths.)

(b) Since BC = 3AD, and AD is parallel to BC, triangle AED and triangle BEC must be similar triangles. As triangle AED and triangle BEC are similar triangles, all sides of triangle BEC must be 3 times the similar sides of triangle AED, hence EC = 3AE.


P.S. Credits to ForbiddenSinner for the solutions and explanations.

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