The side AB of a parallelogram ABCD is produced to X and the bisector of angle CBX meets DA produced and DC produced to E and F respectively. Prove that DE=DF=BA+BC
Please answer fast.

Question

The side AB of a parallelogram ABCD is produced to X and the
bisector of angle CBX meets DA produced and DC produced to E and F
respectively Prove that DE=DF=BA+BC
Please answer fast

Abhishek Kr. Jha · Accepted Answer

Here's the link for the solution of your query:

https://www meritnation
com/ask-answer/question/the-side-ab-of-a-parallelogram-is-produced-to-x-and-the-bise/quadrilaterals/6877958

The side AB of a parallelogram ABCD is produced to X and the bisector of angle CBX meets DA produced and DC produced to E and F respectively. Prove that DE=DF=BA+BCPlease answer fast.

The side AB of a parallelogram ABCD is produced to X and the bisector of angle CBX meets DA produced and DC produced to E and F respectively. Prove that DE=DF=BA+BC
Please answer fast.