In Triangle` A B C` with fixed length of `B C ,` the internal bisector of angle `C` meets the side `A B a tD` and the circumcircle at `E` . The maximum value of `C D×D E` is `c^2` (b) `(c^2)/2` (c) `(c^2)/4` (d) none of these

Options

A. `(b^(2))/4`
B. `(c^(2))/4`
C. `(a^(2))/4`
D. none of these

Explanation

Since CD is bisector of `angleC`.
`therefore(BD)/(AD)=a/b`
`rArr(AD)/(BD)=b/a`
`rArr(AD+BD)/(BD)=(b+a)/arArrC/(BD)=(a+b)/arArrBD=(ac)/(a+b)`
`AD=(bc)/(a+b)`

Now,
`CD.DE=AD.DB`
`rArrCD.DE=(abc^(2))/((a+b)^(2))`
`rArrCD.DEle(c^(2))/4` `[because(a+b)/2gesqrt(ab)rArr(ab)/((a+b)^(2))le1/4]`