# CONVERSE OF BASIC PROPORTIONALITY THEOREM EXAMPLES

Example 1 :

In the figure given below, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR. Solution :

In triangle OPQ,

Given that :

AB is parallel to PQ

(OA/AP)  =  (OB/BQ) ---(1)

In triangle OPR,

Given that :

AC is parallel to PR

(OA/AP)  =  (OC/CR) ---(2)

(1)  =  (2)

(OB/BQ)   =  (OC/CR)

Hence the sides BC and QR are parallel.

Example 2 :

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO  =  OC/OD.

Solution : Draw OM parallel to AB meeting BC at M.

In triangle ACB,

Given that :

OM is parallel to AB

(OC/OA)  =  (CM/MB) ----(1)

In triangle BDC,

OM is parallel to CD

(BM/MC)  =  (OB/OD)

By taking reciprocals on both sides, we get

(CM/MB)  =  (OD/OB) ----(2)

(OC/OA)  = (OD/OB)

(OC/OD)  =  (OA/OB)

Hence proved.

Example 3 :

The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO  =  CO/OD. Show that ABCD is a trapezium

Solution : Also, E is the mid-point of AC (Given)

AE  =  EC

AE/EC  =  1 [From equation (i)]

From equation (i) and (ii), we get

By using the converse of basic proportionality theorem, the sides DE and BC are parallel.

Example 4 :

ABCD is a trapezium in which AB ∥ DC and its diagonals intersect each other at the point O. Show that (AO/BO) = (CO/DO) In ΔADC, we have OE || DC

AE/ED = AO/CO  ...(i)

In ΔABD, we have OE || AB

DE/EA = DO/BO ...(ii)

From equation (i) and (ii), we get

AO/CO  =  BO/DO

AO/BO  =  CO/DO Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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