In ` A B C ,` the bisector of the angle A meets the side BC at D andthe circumscribed circle at E. Prove that `D E=(a^2secA/2)/(2(b+c))`

Explanation

AD = length of angle bisector `= (2b cos.(A)/(2))/(b + c)`

we have `ck + bk = a`
`k = (a)/(b + c)`
From secant property of circle, we have
`AD xx DE = BD xx DC`
`rArr AD xx DE = b c k^(2)`
`rArr (2bc cos.(A)/(2))/(b + c) xx DE = (bc a^(2))/((b+c)^(2))`
Hence, `DE = (a^(2) sec.(A)/(2))/(2(b + c))`