Triangulation of parallelogram

Geometry Level pending

A B C D ABCD is a parallelogram. E E is a point on line segment B C BC , and F F is the intersection point of B D BD and E A EA . If [ A F D ] = 50 [ AFD] = 50 and [ B F E ] = 18 [BFE] = 18 , what is [ A B C D ] [ABCD] ?

Details and assumptions

[ P Q R S ] [PQRS] denotes the area of figure P Q R S PQRS .


The answer is 160.

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Calvin Lin by Calvin Lin

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1 solution

Calvin Lin Staff
May 13, 2014

Let point G G and H H be on A B AB and C D CD respectively such that G F H GFH is a straight line and G H ∥ B C GH \parallel BC .

F A D FAD and F E B FEB are similar triangles, hence B E A D = [ F B E ] [ F A D ] \frac{ BE} {AD} = \sqrt{ \frac{ [FBE] } { [FAD] } } . Thus [ B F E ] [ B F C ] = B E B C = B E A D = [ F B E ] [ F A D ] = 3 5 \frac{ [BFE]} { [BFC] } = \frac{BE}{BC} = \frac{ BE}{AD} = \sqrt{ \frac{ [FBE] } { [FAD] } } = \frac{3}{5} . This gives [ B F C ] = 5 3 [ B F E ] = 30 [BFC] = \frac{5}{3} [BFE] = 30 .

Hence [ A B C D ] = [ A G H D ] + [ B G H C ] = 2 [ A F D ] + 2 [ F B C ] = 100 + 60 = 160 [ABCD] = [ AGHD] + [BGHC] = 2 [AFD] + 2 [FBC] = 100 + 60 = 160 .

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