Question

pls solve this ..
pls mates​

Answer 1

Step-by-step explanation:

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