Theorem 12

Transcript

Theorem 12 If two opposite angles of a quadrilateral add up to 180°, then the vertices of the quadrilateral lie on a circle, i.e., they are concyclic. Given: ABCD is a quadrilateral such that ∠BAC + ∠BDC = 180° ∠ABD + ∠DCA = 180° To Prove: ABCD is a cyclic quadrilateral Proof: Since A, B, C are non–collinear One circle passes through three collinear points Let us draw a circle C1 with centre at O Let us assume D does not lie on C1 Let circle intersect AD at D’ Since ABCD’ is a cyclic quadrilateral And, we know that Opposite angles in a cyclic quadrilateral are Supplementary Thus, ∠ BAC + ∠BD’C = 180° But given that ∠BAC + ∠BDC = 180° Comparing (1) & (2) ∠BD’C = ∠BDC In ∆ BDD’ By Exterior Angle property Which is exterior angle is sum of interior opposite angles ∠ AD’B = ∠BDD’ + ∠D’BD Putting ∠BDD’ = ∠ADB ∠ AD’B = ∠ADB + ∠D’BD Putting ∠ AD’B = ∠ ADB from (3) ∠ ADB = ∠ADB + ∠D’BD ∠ ADB – ∠ADB = ∠D’BD 0 = ∠D’BD ∠D’BD = 0 This means that D’ and D coincide Thus, our assumption was wrong So, Point D lies on circle ∴ ABCD is a cyclic quadrilateral Hence proved