Question

we are given a triangle ABC in which a line is parallel to the side BC intersect other two side A B and ac at d and E is respectvely ​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

ANSWER

Given: In a △PQR, line l∥ side QR, line l intersect the sides PQ and PR in two distinct points M and N respectively.

To prove:

MQ

PM

=

NR

PN

... (i)

Construction: segQN and segRM are drawn.

Proof:

A(△QMN)

A(△PMN)

=

MQ

PM

(Both triangles have equal height with common vertex M)

∴

A(△RMN)

A(△PMN)

=

NR

PN

... (ii)

But A(△QMN)=A(△RMN), because they are between parallel lines MN and QR and have equal height corresponding to their common base MN ..... (iii)

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

A(△QMN)

A(△PMN)

=

A(△RMN)

A(△PMN)

∴

MQ

PM

=

NR

PN

[henceproved]