CE and DE are equal chords of a cricule with centre O. if `angleAOB=90^(@)`find `ar(triangleCED) ar(triangleAOB)`

Explanation

Since O is the centre of circle.
`angleE=90^(@)` (angle ina semicircle is right angle)
`angle1=angle3 (each 45^(@)`
`triangleCED~ triangleAOB` (AA corollary)
`(ar(triangleCED))/(ar(triangleAOB))=(CD^(2))/(AB^(2))` ( area ratio theorem)
= `((2OB)^(2))/(OA^(2) + OB^(2))`
(CD = diameter and OB= radius so, CD=2 , OB and using pythagoras theorem) .
`(4OB^(2))/(OB^(2)+OB^(2))` (OA=OB each radius)
`(4OB^(2))/(2OB^(2))=2/1`
Hence , `(ar(triangleCED))/(ar(triangleAOB))=2/1`