Question

The diagonals `AC` and `BD` of a cyclic quadrilateral `ABCD` interest at right angles at E (figure). A line l drawn through E and perpendicular to AB meets CD at F. Prove that F is the mid-point of CD.

Answer 1

ABCD is a cyclic quadrilateral in which diagonals AC and BD intersect in E at right angles , l is a line through E and perpendicular to AB meets CD in F.
We have `angle AEB=90^@ " "(because "the diagonals are at right angles")`
`rArrangle1_angle2 =90^@`
`angle1 =90^@-angle2` .....(1)
`rArrangle1 =90^@-angle2" "(because EMbotAB)`
Again `angle EMB=90^@`
In `Delta EMB`, we have
`angle2+angle4=90^@`
`rArrangle4=90^@-anlge2` .......(2)
From eqs.(1) and (2) , we get
`angle1=angle4`
But `angle1 =angle3` (vertically opposite angles)
`therefore angle3=angle4`
Also, ` angle4=angle5` (angles in the same segment of a circle are equal)
`rArr angle3=angle5`
`rArrCF=EF` (sides opposite to equal angles of a triangle are equal)
Similarly , `DF=EF`
`rArrCF=DF`
Hence, F is the mid-point of CD.