InterviewSolution
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InterviewSolution
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In the figure ABCD is a trapezium in which side AB is parallel to side DC and E is the midpoint of side AD. If F is a point on the side BC such that the line segment EF is parallel to DC.Prove that Fis the mid-point of BC and `EF =1/2 (AB+DC)` |
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Answer» in `/_ ADC` `(AE)/(ED) = (AM)/(MC)` `M` is the mid point of AC `AB || DC || EF` `EF || DC` `EM || DC` so,`AB || EF` `AB || MF` `F ` is also a mid point of BC as M is the mid point of AD `(AM)/(MD) = (BF)/(FD)` `EM = 1/2 DC` `=> AB || QR` `AB = 1/2 QR` If A&B are midpoints of PQ & PR `E & M` are midpoints `EM || DC & EM= 1/2(DC)` & similarly, `MF = 1/2(AB)` `EM + MF = 1/2[AB + CD]` `EF= 1/2[AB + CD]` hence proved |
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