Math, asked by choudharyshyam1976, 5 hours ago

Activity 3: Draw an angle XAY on your notebook and on ray AX, mark points B_{1}, B_{2}; B_{3}, B_{4} and B such that AB 1 =B 1 B 2 =B 2 B 3 =; B_{3}*B_{4} = B_{4}*B . Similarly, on ray AY, mark points C 1 ,C 2 ,C 3 ,; C_{4} and C such that AC 1 =C 1 C 2 =; C_{2}*C_{3} = C_{3}*C_{4} = C_{4}*C Then join B_{1}*C_{1} and BC (see Fig. 6.11). Note that AB 1 B,B = AC 1 C,C (Each equal to 1 4 You can also see that lines B_{1}*C_{1} and BC are parallel to each other, i.e., B 1 C 1 || F Similarly, by joining B_{2}*C_{2}, B_{3}*C_{3} and B_{4}*C_{4} you can see that: AB 2 B 2 B = AC 2 C 2 C (= 2 3 ) and B_{2}*C_{2} 11 BC AB 3 B 3 B = AC 3 C 3 C (= 3 2 ) and B_{3}*C_{3} 1 BC AB 4 B 4 B = AC 4 C 4 C (= 4 1 ) and B_{4}*C_{4} || BC B_{4}; B_{3}; B_{2}; B_{1} A C_{1}; C_{2}; C_{3}; C_{4} C Y B X Fig. 6.11 (1) (2) (3) (4) From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY. Each time, you will arrive at the same result. Thus, we obtain the following theorem, which is the converse of Theorem 6.1:​

Answers

Answered by Anonymous
1

Answer:

it's true

this theorem is true

you can follow it

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Answered by saniyasayyed1522
2

Step-by-step explanation:

Given : The line l intersects the sides PQ and side PR of ΔPQR in the points M and N respectively such that

MQ

PM

=

NR

PN

and P−M−Q, P−N−R.

To Prove : Line l ∥ Side QR

Proof : Let us consider that line l is not parallel to the side QR. Then there must be another line passing through M which is parallel to the side QR.

Let line MK be that line.

Line MK intersects the side PR at K, (P−K−R)

In ΔPQR, line MK∥ side QR

MQ

PM

=

KR

PK

....(1) (B.P.T.)

But

MQ

PM

=

NR

PN

....(2) (Given)

KR

PK

=

NR

PN

[From (1) and (2)]

KR

PK+KR

=

NR

PN+NR

(P−K−R and P−N−R)

∴ the points K and N are not different.

∴ line MK and line MN coincide

∴ line MN∥ Side QR

Hence, the converse of B.P.T. is proved.

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