Math, asked by choudharyshyam1976, 1 month 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 ​

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Answered by Komal549
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Answer:

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Step-by-step explanation:

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