In the given figure, join AE.
Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral.
`:. angleACD+ angleAED=180^(@)` …(i)
[sum of opposite angles in a cyclic quadrilateral is `180^(@)`]
Now, `angleAEB=90^(@)`...(ii)
We know that, diameter subtends a right angle to the circle.
On adding Eqs. (i) and (ii), we get
`(angleACD+angleAED)+angleAEB=180^(@)+90^(@)=270^(@)`
`rArr angleACD+angleBED=270^(@)`
