In the adjoining figure, `AB` is a diameter of the circle with centre `O`. If `AC` and `BD` are perpendiculars on a line `PD`, and `BD` meets the circle at `E`, prove that `AC` `=` `ED`

Explanation

Join `AE`.
Now, `/_ AEB = 90^(@)` [ angle in a semicircle. ]
But, `BED` being a straight line, we have
`/_ AEB + /_ AED = 180^(@)`
`implies 90^(@) + /_ AED = 180^(@)`
`implies /_ AED = 90^(@)`
Now, `/_ EAC + /_ ACD + /_CDE +/_AED = 360^(@)` [sum of the `/_ s` of quad. EACD is `360^(@)` ]
`implies /_ EAC +90^(@)+90^(@)+90^(@) =360^(@)`
`implies /_ EAC =90^(@)`
Thus, each angle of quad. `EACD` is `90^(@)`.
So, `EACD` is a rectangle.
`: AC= ED` [opposite sides of a rectangle].