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प्रश्न
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.
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उत्तर
Join AC and BD
In ΔACD, G and H are the mid-points of DC and AC respectively.
Therefore, GH || AC and GH = `(1)/(2)"AC"` ...(i)
In ΔABC, E and F are the mid-points of AB and BC respectively.
Therefore, EF || AC and EF = `(1)/(2)"AC"` ...(ii)
From (i) and (ii)
EF || GH and EF = GH = `(1)/(2)"AC"` ...(iii)
Similarly, it can be proved that
EF || GH and EH = GF = `(1)/(2)"BD"` ...(iv)
But AC = BD ...(Diagonals of a square are equal)
Dividing both sides by 2,
`(1)/(2)"BD" = (1)/(2)"AC"`
From (iii) and (iv)
EF = GH = EH = GF
Therefore, EFGH is a parallelogram.
Now in ΔGOH and ΔGOF
OH = OF ...(Diagonals of a parallelogram bisect each other)
OG = O ...(Common)
GH = GF
∴ ΔGOH ≅ ΔGOF
∴ ∠GOH = ∠GOF
Now, ∠GOH + ∠GOF = 180°
⇒ ∠GOH + ∠GOH = 180°
⇒ 2∠GOH = 180°
⇒ ∠GOH = 90°
Therefore, diagonals of parallelogram EFGH bisect each other and are perpendicular to each other.
Thus, EFGH is a square.
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In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
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