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
Answers
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:
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