Question

in the adjoining figure ABCD is a kite in which ab = a d and c b equals to CD if E, F ,G are respectively the midpoints of Ab and CD prove that angle E, F ,G equals to 90 degree and if GH parallel to AC then h bisects CB ​

Answer 1

Answer:

Step-by-step explanation:

and join the mid points of side ab and ad .let the mid points be e and f resp.

NOW ABD is  triangle and by converse of midpoint theoram EF=1/2 BD AND EF//BD. Similarly  in Δ BCD GH=1/2BD AND GH//BD (let g and h be midpoints of side CD AND BC resp.)

NOW, EF=GH AND EF//GH (by euclid's axiom no.1)

so,EFGH IS //grm

JOIN THE another diagonal ac and let the intersection be O

let the intersection of eh and ef with bd and ac be x and y resp.

 

now EX // yo and EY//XO SO, EXOH //grm

 diagonals of kite bisect at 90° so ∠o=∠e=90°

 

∠feh=90° 

 so all angles are 90

Answer 2

Step-by-step explanation:

EF||BD by mid point theorem.

MF||ON

Similarly, FN||ON,

And, the diagonals of a kite intersect at right angles.

So, angle EFG is a right angle as the enclosed figure is forming a rectangle.