Question

In a circle a diameter `AB` is drawn and on one side of it an arc `CD` is marked, which subtends an angle `50^@` at the center. `AC` and `BD` are joined and produced to meet at `E`. Find `/_CED`?

Answer 1

`/_CED=65^@`

Explanation:

Consider the following figure, as described in the question, where we have joined `AD` and `BC` and let `/_BOD=x` and `/_AOC=y`.

As chord `BC` subtends angle `/_BOC` at center and `/_BAE` at the circle, we have `/_BAE=1/2(50+x)=25+x/2`

Similarly for chord `AD`, we have `/_ABE=1/2(50+y)=25+y/2`

Hence `/_CED`, being the third angle of `DeltaABE` is given by

`/_CED=180^@-/_BAE-/_ABE`

= `180^@-25^@-x/2-25^@-y/2=130^@-(x+y)/2`

= `130^@-1/2(180^@-50^@)=130^@-65^@=65^@`