Basic properties of circles - (10.4)

Understand the tests for concyclic points and cyclic quadrilaterals

  • Graph
  • Theory
    if A and D are two points on the same side of the line BC and ∠BAC = ∠BDC , then A , B , C and D are concyclic
    converse of s in the same segment)
  • Examples
    In the figure, AEC and BED are straight lines. It is given that \(\angle BDC = 30^\circ \),
    \(\angle ACB = 50^\circ \) and \(\angle ABC = 100^\circ \). Prove that A, B, C and D are concyclic.
  • Solutions

    In ABC,
    \(\begin{array}{1}\angle BAC + \angle ABC + \angle ACB = 180^\circ \\\angle BAC + 100^\circ + 50^\circ = 180^\circ \\\angle BAC = 30^\circ \end{array}\) ( sum of )
    \(\therefore \angle BAC = \angle BDC = 30^\circ \)
    \(\therefore\) A, B, C and D are concyclic. (converse of s in the same segment)

  • Graph
  • Theory
    if a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic
    opp angles supp
  • Examples
    In the figure, AEC and BED are straight lines. It is given that \(\angle BAD = 82^\circ \),
    \(\angle CBD = 32^\circ \) and \(\angle BDC = 50^\circ \). Prove that A, B, C and D are concyclic.
  • Solutions
    In BCD,
    \(\begin{array}{1}\angle BCD + \angle CBD + \angle BDC = 180^\circ \\\angle BCD + 32^\circ + 50^\circ = 180^\circ \\\angle BCD = 98^\circ \end{array}\) ( sum of )
    \(\begin{array}{1}\angle BAD + \angle BCD = 82^\circ + 98^\circ \\ = 180^\circ \end{array}\)
    \(\therefore\) A, B, C and D are concyclic. (opp. s supp.)
  • Graph
  • Theory
    if the exterior angle of a quadrilateral equals its interior opposite angle, then the quadrilateral is cyclic
    (ext.   int. opp. )
  • Examples
    In the figure, D and E are points on AB and AC respectively. \(\angle DAE = 41^\circ \),
    \(\angle AED = 62^\circ \) and \(\angle ACB = 77^\circ \). Prove that BCED is a cyclic quadrilateral.
  • Solutions
    In ADE,
    \(\begin{array}{1}\angle ADE + \angle DAE + \angle AED = 180^\circ \\\angle ADE + 41^\circ + 62^\circ = 180^\circ \\\angle ADE = 77^\circ \end{array}\) ( sum of )
    \(\therefore \angle ADE = \angle BCE = 77^\circ \)
    \(\therefore\) BCED is a cyclic quadrilateral. (ext.   int. opp. )