Question

50 points please answer this then I will mark as brainliest

Answer 1

given diagonals of ABCD are prependicular therefore it is a Rhombus
In Rhombus all sides are equal
therefore mid points are also equal
in length
In traingle ABC ,Fand H midpoints therefore by mid point theorem FH is parallel to AC and1/2ofAC...(1)
In triangle ADC ,E and G are mid points,by mid point theorem EG is parallel to ACand1/2of AC...(2)
from (1) and (2)..
EG=FH
EG parallel to FH
SIMILARLY FROM TRIANGLES.,DAB and DCB ..---->>EF Parallel to GH and EF=GH
so if a pair of opposite sides are equal and parallel it is a parallelogram
now lets draw diagonals in a parallelogram they bisect each other
let their intersecting point is O in trianglesEGH andEFH
EG=EF
EH=EH
GH=FH
Therefore they are congruent by SSS congruence
<EHF=<EHG by (c.p.c.t.)
similarly in trngle EFG andFGH <EGF=<FGH
asH=E,G=F
G=H as G+H=180
2G=180
G=90 SO EFGH is a rectangle

Answer 2

Answer:

Step-by-step explanation: