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Question

Let ABC be a triangle. Let D,E,F be the points respectively on the segments BC,CA,AB such that AD,BE,CF concur at the point K. Suppose BDDC=BFFA and ADB=AFC. Is ABE=CAD.?

A
True
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B
False
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Solution

The correct option is A True


As It is given ,
BDDC=BFFA and ADB=AFC

The line DF and CA are parallel
We have BDK=ADB=180BFK
So that BDFK is cyclic quadrilateral
FBK=BDK

ABE=FBK=FDK=FDA=DAC
finaly ABE=CAD

1912749_303141_ans_d10698fe8b1a4a3e898eecd8643e0586.png

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