Question 1
ABC is a triangle. Locate a point in the interior of $Δ$ ABC which is equidistant from all the vertices of $Δ$ ABC.

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Solution

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In $Δ$ ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of $Δ$ ABC

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ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

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Construct a $Δ$ ABC, in which BC = 6 cm, AB = 9 cm and $∠$ ABC = 60 $^{\circ}$

i. Construct the locus of all points inside $Δ$ ABC, which are equidistant from B and C.

ii. Construct the locus of the vertices of the triangle with BC as base, which are equal in area to triangle ABC.

iii. Mark the point Q in your construction, which would make $Δ$ QBC equal in area to $Δ$ ABC, and isosceles.

iv. Measure and record the length of CQ.

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