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Question

State and prove the converse of the following theorem. “If a line divides any two sides of a triangle in the same ratio it must be parallel to the third side”. Using the result, prove the following:

"A line drawn from the midpoint of a non-parallel side of a trapezium, parallel to the parallel sides, bisects the other non-parallel side.


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Solution

Step 1: Note the given data and draw a diagram

Let ABC be a triangle.

In ABC, the line DE intersects AB and BCinD and E respectively.

Given that ADDB=AEEC……(i)

Let the line DE is not parallel to BC.

Construction: Draw a line DF ( assume that DF is parallel to BC)

Step 2: Proving the basic proportionality theorem :

In ADFandABC,

Since DFBC, so according to basic proportionality theorem

ADF=ABC(Correspondingangles)AFD=ACB(Correspondingangles)ADF~ABC(by,AAsimilaritycriteria)

Since, ADF~ABC.

we know that, the corresponding sides of similar triangles are proportional.

ADDB=AFFC…..(ii)

Equating equation (i) and equation (ii)

AEEC=AFFC

Adding 1 both sides

AEEC+1=AFFC+1AE+ECEC=AF+FCFCACEC=ACFCEC=FC

This is only possible when E is coincide with F.

Thus DE is parallel to BC.

Hence proved.

Step 3: Construct a trapezium and note the given data

Let ABCD be a trapezium with ADBC.

let, F are midpoints of AD , draw a line from F meeting BC at E. Such that, EFABDC.

Given, AF=FD.

The diagram is shown below.

Step 4: Applying Thales theorem in ACD and CAB

Thales theorem states that if a line drawn parallel to one side of a triangle intersects the other two sides in distinct points, the other two sides are divided in the same ratio.

Here OFCD

AOOC=AFFD............(i)

Here OEAB

AOOC=BEEC............(ii)

Equating equations (i) and (ii)

AFFD=BEECAFAF=BEECAF=FDBE=EC

Hence, the statement “A line drawn from the midpoint of a non-parallel side of a trapezium, parallel to the parallel sides, bisects the other non-parallel side” is proved


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