TRIANGLES SIMILARITY
Asked by gunjansingh | 17th Feb, 2010, 05:55: AM
the figure formed by joining the mid-points of pairs of consecutive sides of a quadrilateral is a parallelogram. Answer: Given: If AC=BD then EFGH is a rhombus.
ABCD is a quadrilateral. E, F, G and H are the mid-points of
AB, BC, CD and DA respectively.
To prove:
EFGH is a parallelogram.
Construction:
Join A to C and D to B.
Proof:
Consider D ACD,
mid-points of two sides of a triangle is parallel to the third side and half of it)
Consider D ACB,
mid-points of two sides of a triangle is parallel to the third side and half of it)
From (i) and (ii),
HG=EF and HG||EF
\ EFGH is a parallelogram.
Answered by | 17th Feb, 2010, 03:36: PM
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