Diagonals AC and BD of a cyclic quadrilateral ABCD intersect at right angles at E as shown in figure.A line drawn through E and perpendicular to AB meets at M and CD meets at F. prove that F is the mid point of CD.
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In triangle EFC&EFD
angle EFC= angle EFD=90°(Given)
EF=FE (Common)
CE=DE(side opposite to equal angle are equal)
By SAS rule ∆EFC=~∆EFD
By cpct DF=FC
OR F is the mid point of DC
angle EFC= angle EFD=90°(Given)
EF=FE (Common)
CE=DE(side opposite to equal angle are equal)
By SAS rule ∆EFC=~∆EFD
By cpct DF=FC
OR F is the mid point of DC
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