Theorem 1

Transcript

Theorem 1 There is a unique circle passing through three non-collinear points Given: P, Q and R are three non–collinear points To Prove: Only one circle passes through PQR Construction: Draw AB perpendicular bisector of PQ at M and CD, Perpendicular bisector of QR at N Proof : Since P,Q, R are non-Collinear So, PQ ∦𝑄𝑅 AB and CD are also not parallel Thus, AB & CD has to intersect at a point Let that point be O Join OP, OQ, OR IN ∆ OPM and ∆ OQM OM = OM ∠ OMP = ∠OMQ PM = QM ∴ ∆ OPM ≅ ∆OQM (Common) (Both 90°, AB ⊥ PQ ) (AB bisects PQ) (SAS rule) Now, by CPCT ∴ OP = OQ Similarly, OQ = OR From (1) and (2) OP = OQ = OR Let OP = OQ = OR = 𝑟 Now, taking O as center and r as radius construct a circle. Hence, circle passes through 3 non-collinear points P, Q, R Now, we have to prove that only one such circle is there Proving only 1 such circle is there Let us assume there’s another circle with centre O’ and radius r’ passing through P, Q, R. Now, drawing a perpendicular bisector from new center O’ Let Perpendicular bisector be O’M’ i.e. O’M’ ⊥ PQ Now, We know that Perpendicular from center to a chord bisects it Thus, PM’ = QM’ So, M’ is the mid point of PQ. ∴ O’ M’ is the perpendicular bisector of PQ. But OM is the perpendicular bisector of PQ As a line can have only one perpendicular bisector, Hence, O’M’ coincides with OM O’ coincides with O Thus, there is only 1 circle which passes through 3 points P, Q, R Hence proved