Q11 of 24 Page 157

In the trapezium ABCD, AB || DC and AB > DC; the midpoints of two diagonals AC and BD are E and F respectively. Let us prove that, .


As E and F are the midpoints of sides AC and BD respectively, therefore by the theorem:-


The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,


⇒ EF || CD


Now, we have extended EF to H, so in ΔACD, since E is midpoint of AC and EH||CD, so by applying theorem:-


Through the mid-point of any side, if a line segment is drawn parallel to second side, then it will bisect the third side and the line segment intercepted by the two sides of the triangle is equal to half of the second side.



Similarly, applying above theorem in ΔADB as F is midpoint of BD and FH||AB,



As FE = FH – EH,


So by substituting above values, we get,



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