Construct a triangle similar to a given triangle ABC such that each of its sides is two thirds of the corresponding sides of triangle ABC. It is given that BC= 6 cm, angle B = 50^{o}and angle C = 60^{o}.

### Asked by Topperlearning User | 27th Jul, 2017, 01:45: PM

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Steps of construction;

Steps of construction;

i. Draw BC = 6cm

ii. At B construct angle CBX = 50^{o}

iii. At C construct angle BCY = 60^{o}

iv. Let BX and CY intersect at A. Triangle ABC thus obtained is the given triangle.

v. Construct an acute angle CBZ at B on opposite side of the vertex A of triangle ABC.

vi. Mark off 3 (greater 3 in 2/3) points B_{1, }B_{2, }B_{3} on BZ such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}.

vii. Join B_{3 }to C and draw a line through B_{2} parallel to B_{3}C, intersecting BC at P.

viii. Through P draw PQ parallel to AC, meeting AB at Q

ix. BPQ is the required triangle, each of whose sides is two-thirds of the corresponding sides of triangle ABC.

### Answered by | 27th Jul, 2017, 03:45: PM

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