Page 3 - ws2
P. 3
Sol.
Given:
Trapezium ABCD in which AB||DC. M and N are the mid-points of the diagonals AC and
BD respectively.
To Prove:
(i) MN||AB||DC
(ii) MN = (AB - CD)
Construction:
Join CN and produce it to meet AB at E.
Proof:
In triangles CDN and EBN,
DCN = BEN [alternate angles]
CDN = EBN [alternate angles]
DN = NB [N is the mid-point of BD]
Therefore, CDN EBN [AAS]
CD = EB and CN = NE [corresponding parts of congruent triangles]
Therefore, in CAE, points M and N are the mid-points of AC and EC respectively.
(1/2 mark)
Therefore, MN||AE or MN||AB
and MN = AE [by mid point theorem]
= (AB - EB)
= (AB - CD) [ EB = CD]
Hence, MN||AB||CD and MN
= (AB - CD)
