​abc is a triangle inscribed in a circle. the bisector of angle bac meets bc at d and being produced meets the circle at e . prove that angle ace=angle cde 

Dear Student,

Please find below the solution to the asked query :

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{prove} \mathrm{that} \\ \angle \mathrm{ACE}= \angle \mathrm{CDE} \\ \mathrm{proof}- \\ \angle \mathrm{CDE}= \angle \mathrm{DBE}+\angle \mathrm{BED} \\ = \angle \mathrm{CAE}+\angle \mathrm{ACB} \left[\mathrm{Since} \angle \mathrm{DBE} \mathrm{and} \angle \mathrm{CAE} \mathrm{are} \mathrm{on} \mathrm{same} \mathrm{base}\right] \\ = \angle \mathrm{BAE}+\angle \mathrm{ACB} \left[\mathrm{Since} \angle \mathrm{BAE} \mathrm{and} \angle \mathrm{CAE} \mathrm{are} \mathrm{on} \mathrm{same} \mathrm{base}\right] \\ = \angle \mathrm{ACE} \end{gathered}$$

Hope this would clear your doubt about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards