Question

37.
If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, prove that the other two sides are divided in
the same ratio.
​

Answer 1

Answer:

Step-by-step explanation:

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.

[ Basic Proportionality Theorem Or Thales Theorem ]

Given:

In ∆ABC , which intersects AB and AC at D and F respectively.

RTP:

AD/DB = AE/EC

Construction:

Join B , E and C ,D and then draw

DM ⊥ AC and EN ⊥ AB.

Proof:

Area of ∆ADE = 1/2 × AD × EN

Area of ∆BDE = 1/2 × BD × EN

So, A(∆ADE)/A(∆BDE)

= 1/2 × AD × EN / 1/2 × BD × EN

= AD × BD ------------(1)

Again Area of ∆ADE = 1/2 × AE × DM

Area of ∆CDE = 1/2 × EC × DM

So, A(∆ADE)/A(∆CDE)

= 1/2 × AE × DM / 1/2 × EC × DM

= AE/EC--------------(2)

Observe that ∆BDE and ∆CDE are on the same base DE and between same parallels BC and DE.

So, A(∆BDE) = A(∆CDE) ------------(3)

From (1),(2) & (3),. we get

= AD/DB = AE/EC

Hence , proved .

Answer 2

Given: ΔPQR, in which XY || QR, XY intersects PQ and PR at X and Y respectively.

To prove:

Construction: Join RX and QY and draw YN perpendicular to PQ and XM perpendicular to PR.

Proof:

Since, Area of a triangle =

Therefore, ar (ΔPXY) = …(1)

Also, ar (ΔPXY) = …(2)

Similarly, ar (ΔQXY) = …(3)

And, ar (ΔRXY) = …(4)

Dividing (1) by (3), we get

Again, dividing (2) by (4), we get

Since the area of triangles with same base and between same parallel lines are equal, so

(as and are on same base XY and between same parallel lines XY and QR)

Therefore, from (5), (6) and (7), we get

Hence, proved.