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In the adjoining figure, ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively, then prove that MN || AB.

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Given: ABCD is a trapezium. AB || DC. 

Points M and N are midpoints of diagonals AC and DB respectively. 

To prove: MN || AB 

Construction: Join D and M. Extend seg DM to meet seg AB at point E such that A-E-B. 

Proof: 

seg AB || seg DC and seg AC is their transversal. [Given] 

∴ ∠CAB ≅ ∠ACD [Alternate angles] 

∴ ∠MAE ≅ ∠MCD ….(i) [C-M-A, A-E-B] 

In ∆AME and ∆CMD,

∠AME ≅ ∠CMD [Vertically opposite angles] 

seg AM ≅ seg CM [M is the midpoint of seg AC] 

∠MAE ≅∠MCD [From (i)] 

∴ ∆AME ≅ ∆CMD [ASA test] 

∴ seg ME ≅ seg MD [c.s.c.t] 

∴ Point M is the midpoint of seg DE. … (ii) 

In ∆DEB,

Points M and N are the midpoints of seg DE and seg DB respectively. [Given and from (ii)] 

∴ seg MN || seg EB [Midpoint theorem] 

∴ seg MN || seg AB [A-E-B]

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