# Theorem 10.12

Last updated at April 19, 2017 by Teachoo

Last updated at April 19, 2017 by Teachoo

Transcript

Theorem 10.12 If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic. Given : ABCD is 𝑎 quadrilateral such that ∠BAC + ∠BDC = 180° 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 Now, ∴ ABCD’ is 𝑎 cyclic quadrilateral ∴ ∠ BAC + ∠BD’C = 180° But given ∠BAC + ∠BDC = 180° Thus, ∠BD’C = ∠BDC Now, In Δ BDD’ ∠BD’C = ∠BDD’ + ∠DBD’ ∠BD’C = ∠BDC + ∠DBD’ ∠BDC = ∠BDC + ∠DBD’ ∠BDC – ∠BDC = ∠DBD’ ∴ ∠DBD’ = 0 ∴ D’ and D Coincides ∴Our assumption was wrong ⇒ Point D lies on circle C1 ∴ A, B, C, D are concyclic. Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.