Question

OACB is a parallelogram. OA = a and OB = b. The points M, S, N and T divide OB, BC, CA and AO in the same ratio respectively. The lines ST and MN intersect at the point D. Show that the lines MN and ST bisect one another?

Answer 1

Given

OACB is a parallelogram. OA = a and OB = b. The points M, S, N and T divide OB, BC, CA and AO in the same ratio ( say `m:n` ) respectively. The lines ST and MN intersect at the point D.

Rtp

D is the mid point of MN and ST

Proof

OACB is a parallelogram. So `BC=OA=a and AC=OB=b`

Let `m/n=(OM)/(MB)=(BS)/(SC)=(CN)/(NC)=(AT)/(OT)`

In `DeltaTOM and Delta NCS`

So `OM=CN=(bm)/(m+n) and OT =CS=(an)/( m+n)`

and `angle MOT=angleNCS->"opposite angles of a paralleogram"`

Hence `DeltaTOM ~= Delta NCS`

This means `MT~=NS`

similarly `DeltaTAN ~= Delta MBS`

This means `NT~=MS`

So opposite sides of the quadrilateral MSNT are congruent.

Hence the quadrilateral must be parallelogram. This implies that the point of intersection of D of the its diagonals `MN and ST` must be the mid point of each. This means they bisect each other.