Question

ABCD is a parallelogram.E is the mid point of AB and F is the mid point of DC. GH is any straight line that intersects AD,EF, and BC at G,I,H respectively. Prove that GI=HI.
The ch is mid point theorem
don't give any wrong procedure​

Answer 1

is 400m)

2x+26=200

2x=174

x=87m

Length =87m and as stated Breadth =87+26=113mLet the length of the park be x

Then according to the question,

Breadth=x+26 and 2(x+x+26)=400 (perimeter is 400m)

2x+26=200

2x=174,

Breadth=x+26 and 2(x+x+26)=400 (perimeter is 400m) (perimeter is 400m)

2x+26=200

2x=174

Answer 2

Answer:

Construction: Draw a line parallel to AB through I intersecting AD at X and BC at Y

Consider triangles XIG and YIH

Angle XIG= Angle YIH (vertically opposite angles)

Angle IGX =Angle IHY (Alternate angles between parallel lines AD and BC)

AEIX and EBIY are two parallelograms thus formed

where

AE= XI=EB=IY (E is midpoint of AB)

Hence a side and 2 angles from each triangle are congruent

Hence by SAA test of congruency GIX and HIY are congruent triangles

Hence GI=HI