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Question

In trapezium ABCD, AB||DC. E and F are points on non-parallel sides AD and BC respectively such that EFAB. Prove that AEED=BFFC.
1036109_8d00f42ae8ad4c9b81bcf417d9ed52df.png

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Solution

ABDC & EFDC, therefore
ABEFDC
Join AC which intersects EF at G. In ADC ,
EGDC [EF is the extension of EG]
AEED=AGGC (1) [Lines drawn parallel to one side of triangle intersects the other two sides in distinct points. Then it is divided the other two sides in same ratio]
Similarly in ABC , ABGF , Therefore
BFFC=AGGC (2)
From (1) & (2) ,
AEED=BFFC

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