Question

plzz answer question no 5,1st one​

Answer 1

F, E are midpoints of AC and BC respectively.

We know that the line segment joining the midpoints of two sides of a triangle is parallel to the third side, and its length is half of that of the third side.

∴ FE ║ AB

∴ ∠ABC = ∠FEC                        (corresponding)

Let it be x.

So ∠FEB becomes 180 - x, as ∠FEB and ∠FEC are linear pairs.

Consider ΔABP.

It's a right triangle and D is the midpoint of the hypotenuse AB.

So D is the circumcenter of ΔABP, because the midpoint of the hypotenuse of any right triangle is its circumcenter.

∴AD = BD = PD are the circumradii of ΔABP.

Consider ΔBDP.

Found that BD = PD.

∴ ∠BPD = ∠PBD = x                       (∵ ∠PBD = ∠ABC)

∴ ∠DPC = 180 - x                            (∠BPD, ∠DPC are linear pairs)

D, F are midpoints of AB and AC respectively.

∴ DF ║ BC

∴ ∠BPD = ∠PDF = x                        (alternate)

So, in quadrilateral DFEP,

We got the following:

∠FEP = ∠DPE = 180 - x    and    ∠PDF = x.

∠FEP + ∠PDF = 180 - x + x = 180

∴ ∠FEP, ∠PDF are supplementary.

In any quadrilateral, if one pair of opposite sides is supplementary, then so is the other.

∵ Here, ∠DPE + ∠DFE = 360 - (∠FEP + ∠PDF) = 360 - 180 = 180.

∴ Quadrilateral DFEP is cyclic.

∴ Points D, F, E and P are concyclic.

Hence, proved!

Hope this helps. ^_^

Thank you. :-))