Question

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.

Answer 1

i) If the 4th point has any specific relation with the given 3 points, then subject to that only the construction process can be outlined. However one such point which lie on the circle passing through the given 3 given points is described below.

ii) Let the 3 non collinear points be A, B & C.

iii) Join A-B and B-C.

iv) Using general method of using compass and ruler, draw at A & C perpendiculars to AB and BC respectively. Let these two perpendiculars intersect at D.
Then D is the 4th point lying on the circle passing through A, B & C.

Justification:

i) From this construction ABCD is a quadrilateral.

ii) In this by construction, <BAD = <BCD = 90 deg
==> Some of the opposite angles = 180 deg
So sum of the remaining pair of opposite angles is also 180 deg
Hence ABCD is a cyclic quadrilateral.

Thus D is a point on the circle passing through A, B & C

EDIT:

Let me give one more method also for the same.

i) Let A, B & C be the given 3 non collinear points.
Join AB, BC & CA to make the triangle ABC.

ii) With BC as base construct another triangle BCD congruent to BCA.
[Measure BA; with C as center mark an arc equal to BA; Similarly with B as center mark an arc equal to CA;let these two arcs intersect at D; so we get another triangle BCD]

iii) By construction CD = BA and BD = CA
Also BC = CB [Common base]
Hence, ΔBAC ≅ ΔCDB [SSS congruence axiom]

==> <BAC = <CDB [Corresponding parts of congruence triangles are equal]

==> Angles subtended by the same line are equal.

So chord BC subtends equal angles at the segment; hence D is a point on the circle passing through A, B & C.