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प्रश्न
In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
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उत्तर
Given: `square`ABCD is a trapezium. AB || DC
Points M and N are the midpoints of diagonals AC and DB respectively.
To prove: MN || AB
Construction: Draw line DM which intersects side AB at point T.
Proof:
side DC || side AB …(Given)
And seg AC is a transversal line.
∴ ∠DAC ≅ ∠BAC ...(alternate angles)
∴ ∠DCM ≅ ∠TAM ...(i) ...(A-M-C and A-T-B)
In ∆DCM and ∆TAM,
∠DCM ≅ ∠TAM ...[From (i)]
seg MC ≅ seg MA ...(Point M is the midpoint of seg AC.)
∠DCM ≅ ∠TAM ...(Vertically opposite angles)
∴ ∆DCM ≅ ∆TAM ...(ASA test)
seg DM ≅ seg MT ...(c.s.c.t) ...(ii)
In ∆DTB,
Point N is the midpoint of line DB. ...(Given)
Point M is the midpoint of line DT. ...[From (ii)]
∴ seg MN || side TB ...(Midpoint Theorem)
∴ seg MN || seg AB ...(A-T-B)
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