Question

state and prove basic proportionality theorem​

Answer 1

Hint: To prove this theorem first we will join BE and CD. Then draw a line EL perpendicular to AB and line DM perpendicular to AC. Now we will find the ratio of area of Δ

ADE to Δ

DBE and ratio of area of Δ

ADE to Δ

ECD. Comparing the ratios we will get the final answer.

Complete step-by-step answer:

Now, ΔDBE

and ΔECD

being on the same base DE and between the same parallels DE and BC, we have,ar(ΔDBE)=ar(ΔECD)

then we say that the basic proportionality theorem is proved.

Basic proportionality theorem:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Given:

A ΔABC

in which DE∥BC

and DE intersects AB and AC at D and E respectively.

To prove that:

ADDB=AEEC

Construction:

Join BE and CD.

Draw EL⊥AB

and DM⊥AC

Proof:

We have the

ar(ΔADE)=12×AD×EL

ar(ΔDBE)=12×DB×EL

Therefore the ratio of these two is ar(ΔADE)ar(ΔDBE)=ADDB

. . . . . . . . . . . . . . (1)

Similarly,

ar(ΔADE)=ar(ΔADE)=12×AE×DM

ar(ΔECD)=12×EC×DM

Therefore the ratio of these two is ar(ΔADE)ar(ΔECD)=AEEC

. . . . . . . . . . . .. . . (2)

Now, ΔDBE

and ΔECD

being on the same base DE and between the same parallels DE and BC, we have,

ar(ΔDBE)=ar(ΔECD)

. . . . . . . . . . . (3)

From equations 1, 2, 3 we can conclude that

ADDB=AEEC

Hence we can say that the basic proportionality theorem is proved.

Note: The formula for area of the triangle is given by 12×b×h

where b, h are base and height respectively. If two triangles are on the same base and between the same parallels then the area of those two triangles are equal.