Question

can u answer the above question​

Answer 1

D & F are midpoints

DF ll BC ....midpoint theorem

DF= 1/2 × BC

DF = BE

quadrilateral DBEF is parallelogram

angle B congruent to angle F ..... opposite sides of parallelogram (1)

In triangle APB ,

1. triangle APB is a right angled triangle

2 . D is midpoint of AB

3. DP is median

DP = BD .... (2)

isosceles triangle

angle B = angle DPB

angle F = angle DPB .... from (1)

angle DPB is external angle of quadrilateral DPEF which is congruent to opposite adjacent angles

i.e. angle F

therefore quadrilateral DPEF is cyclic quadrilateral

hope this answer helps you