Question

in quadrilateral ABCD side AB Congress to site ad bisector of angle BAC cards side BC at P and bisector off the AC cut side BC at F prove that seg
EF parallel to seg BD .​

Answer 1

$\bf\large\underline\pink{Solution:-}$

Step-by-step explanation:In ∆ABC,seg AE bisects the angle BAC

AC/AB=CE/BE......(1)..(property of angle bisector)

In ∆CAD ,seg AF bisects angle CAD

AD/AC=DF/FC....(2)...(angle bisector property)

Adding (1)and(2)

CE/BE=DF/FC ...(3)

In∆CDB

CE/BE=CF/FD

seg EF ||seg BD......(converse of basic proportionality theorem)