Q. 74.3( 4 Votes )
Let us prove that
Answer :
In ΔACD, as E and H are the midpoints of AC and CD respectively.
So by applying the theorem:-
The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get,
and EH || AD ……… (1)
Also, by using above theorem in ΔABD, as F and G are midpoints of AB and BD respectively,
and FG || AD ……… (2)
From equations (1) and (2) we see that, EH = FG and EH || AD.
Similarly for ΔBCD, we have G and H as midpoints of BD and CD respectively, by applying above theorem, we get,
and GH || BC ……… (3)
And also in ΔABC,
and EF || BC ……… (4)
From equations (3) and (4) we see that, GH = EF and GH || EF.
So, EHGF is a parallelogram and as IH lie on EH and HJ lie on HG.
⇒ IHJO is also a parallelogram.
As we know that, the angle subtended by the diagonals in a triangle is equal to 90°, and also in a parallelogram, the opposite angles are equal.
⇒ ∠EHG = ∠IOJ = 90°
Hence, quadrilateral HGFE is a rectangle
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In the triangle A
West Bengal Mathematics<span lang="EN-US
West Bengal MathematicsIn the triangle A
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West Bengal MathematicsLet us prove that
West Bengal MathematicsD and E lie on AB
West Bengal MathematicsIn the triangle A
West Bengal MathematicsIn the triangle A
West Bengal MathematicsLet us prove that
West Bengal MathematicsLet us prove that
West Bengal Mathematics