The given figure shows a quadrilateral ABCD in which E,F,G and H are the mid-points of consecutive sides of ABCD. Again P,Q,R and S are the mid-points of the consecutive sides of quadrilateral EFGH. If EFGH is a rectangle, show that : PQRS is a rhombus.
Answers
Answer:
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
(i) In △DAC , S is the mid point of DA and R is the mid point of DC. Therefore, SR∥AC and SR=
2
1
AC.By mid-point theorem.
(ii) In △BAC , P is the mid point of AB and Q is the mid point of BC. Therefore, PQ∥AC and PQ=
2
1
AC.By mid-point theorem. But from (i) SR=
2
1
AC therefore PQ=SR
(iii) PQ∥AC & SR∥AC therefore PQ∥SR and PQ=SR. Hence, a quadrilateral with opposite sides equal and paralle is a parallelogram. Therefore PQRS is a parallelogram.
Step-by-step explanation:
ANSWER
Given: ABCD is a quadrilateral.
Points E,F,G,H are the midpoints of AB,BC,CD and DA.
Draw diagonals AC and BD in the quadrilateral ABCD
Segment HG is the midpoint segment in △ACD
∴ segment HG is parallel to the side AC of the
△ACD (Line segment joining midpoints of two sides of a triangle property.)
Similarly, Segment EF is the midpoint segment in △ABC
∴ Segments EF is parallel to side AC of △ABC.
Since, Segment HG and EF are both parallel to the diagonal AC, they are parallel to each other.
Segment GF is the midpoint segment in △DCB.
∴ Segment GF is parallel to side DB of △ABD.
Segment HE is midpoint segment in △ABD
∴ Segment HE is parallel to side DB of triangle ABD
Since,Segment GF and HE are both parallel to diagonal DB, they are parallel to each other.
Thus, we have proved that in quadrilateral EFGH the opposite sides
HG and EF,HE and GF are parallel by pairs.
Hence, the quadrilateral EFGH is the parallelogram.
solution
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