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Question

To construct a triangle similar to given ABC with its sides 23 of that of ΔABC, locate points on ray BX at equal distances as B1,B2,B3,.... such that CBX is acute. The points to be joined in the next step are:

A
B4,C
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B
B3,C
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C
B1,C
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D
B2,C
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Solution

The correct option is B B3,C
PQB is the required triangle.
Since side BQ is 23 times side BC.
BQ=23×(BQ+CQ)3BQ=2BQ+2CQBQ=2CQCQBQ=12
Therefore, Q divides BC in ratio 2:1.
So, point B3 should be connected to C, and B2Q should be drawn parallel to B3C.

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