In the given figure, two chordsAB and CD of a circle AYDZBWCX intersect at right angles. Prove that
arc CXA `+` arc DZB `=` arc AYD `+` arc BWC `=` semicircle.

Explanation

GIVEN Two chords AB and CD of a circle AYDZBWCX, intersecting at a point such that `AB _|_ CD`.
CONSTRUCTION Let O be the centre of the given circle. Join OA, OB,OC,OD and BC.
PROOF In `Delta EBC`, we have
`/_ EBC +/_ECB+/_CEB= 180^(@)`
`implies /_ EBC +/_ECB = 180^(@)-/_CEB= (180^(@)-90^(@))= 90^(@)`
`implies /_ ABC +/_DCB = 90^(@)` .....(i)
Since the angle subtended by an arc of a circle at its centre is double the angle subtended by it any point of the remaining part of the circle, we have
`/_ AOC = 2 /_ ABC` and `/_ DOB = 2 /_ DCB`
`implies /_ AOC + /_ DOB = 2 ( /_ ABC + /_ DCB ) = ( 2 xx 90^(@)) = 180^(@) ` [ using (i) ]
`implies ` arc CXA `+` arc DZB `=` semicircle.
Semilarly, arc AYD `+` arc BWC `=` semicircle.