The diagonals of a cyclic quadrliateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.
The diagonals of a cyclic quadrliateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.
Let ABCD be a cyclic quadrilateral such that its diagonals AC and BD intersect in O at right angles.
Let OL ⊥ AB such that LO produced meet meet CD in M.
To prove: M bisects CD i.e. CM = MD.
Consider arc AD,
It makes angles ∠MCO and ∠LBO in the same segment.
∴ ∠MCO = ∠LBO ........ (1)
In ∆OLB right angled at L,
∠LBO + ∠BOL + ∠OLB = 180°
⇒ ∠LBO + ∠BOL + 90° = 180°
⇒ ∠LBO + ∠BOL + 90° ........ (2)
Since, LOM is a straight line,
∴ ∠LOB + ∠BOC + ∠COM = 180°
⇒ ∠LOB + 90° + ∠COM = 180°
⇒ ∠LOB + ∠COM = 90° ............. (3)
From (2) and (3)
⇒ ∠LBO + ∠BOL = ∠LOB + ∠COM
⇒ ∠LBO = ∠COM ............ (4)
From (1) and (4),
∠MCO = ∠COM
⇒ MO = MC
Similarly, MO = DM.
Hence, MC = DM
Let ABCD be a cyclic quadrilateral such that its diagonals AC and BD intersect in O at right angles.
Let OL ⊥ AB such that LO produced meet meet CD in M.
To prove: M bisects CD i.e. CM = MD.
Consider arc AD,
It makes angles ∠MCO and ∠LBO in the same segment.
∴ ∠MCO = ∠LBO ........ (1)
In ∆OLB right angled at L,
∠LBO + ∠BOL + ∠OLB = 180°
⇒ ∠LBO + ∠BOL + 90° = 180°
⇒ ∠LBO + ∠BOL + 90° ........ (2)
Since, LOM is a straight line,
∴ ∠LOB + ∠BOC + ∠COM = 180°
⇒ ∠LOB + 90° + ∠COM = 180°
⇒ ∠LOB + ∠COM = 90° ............. (3)
From (2) and (3)
⇒ ∠LBO + ∠BOL = ∠LOB + ∠COM
⇒ ∠LBO = ∠COM ............ (4)
From (1) and (4),
∠MCO = ∠COM
⇒ MO = MC
Similarly, MO = DM.
Hence, MC = DM
