Question

9.
D
С
M
w
In the adjacent figure 5.44, ABCD is a
trapezium. AB || DC. Points M and N are
midpoints of diagonal AC and DB
respectively then prove that MN || AB.
A
B
Fig. 5.44​

Answer 1

$\huge\underline{\overline{\mid{\bold{\color{maroon}{\mathcal{Question:-}}\mid}}}}$

ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonal AC and DB

respectively then prove that MN || AB.

$\huge\bf\purple{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

$\sf\small\underline\red{Given:-}$

ABCD is a trapezium.

AB || DC.

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

$\sf\small\underline\red{To \: prove:-}$

MN || AB

$\sf\small\underline\red{Construction:-}$

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

$\sf\small\underline\red{Proof:-}$

segment 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]