The proof that the perpendicular from the point of their intersection on any side when produced backward bisects the opposite side is;
Proven below
Angles in a cyclic quadrilateral
I have attached an image showing the given cyclic quadrilateral.
We see from the attached cyclic quadrilateral that diagonal AC and diagonal BD intersect at point O and they form at right angles.
Now, we also see that length OL is perpendicular to length AB such that length LO when extended, meets length CD at point M.
Looking at arc AD, we see that ∠MCO and ∠LBO are produced in the same segment and as such they will be equal and we can write that;
∠MCO = ∠LBO ........(eq 1)
Also, we see that ∆OLB is a right angled triangle since L is perpendicular to AB as previously stated. Since sum of angles in a triangle is 180°, then we can write that;
∠LBO + ∠BOL + ∠OLB = 180°
Thus;
∠LBO + ∠BOL + 90° = 180°
⇒ ∠LBO + ∠BOL = 90° ........(eq 2)
Since, since L is both perpendicular to AB and CD, then we can say that LOM is a straight line. Thus;
∠LOB + ∠BOC + ∠COM = 180°
⇒ ∠LOB + 90° + ∠COM = 180°
⇒ ∠LOB + ∠COM = 90° .............(eq 3)
Putting ∠LOB + ∠COM for 90° in eq 2 gives us;
∠LBO + ∠BOL = ∠LOB + ∠COM
∠BOL is same as ∠LOB and they will cancel out to give us;
∠LBO = ∠COM ............(eq 4)
From eq 1, we can put ∠MCO for ∠LBO in eq 4;
∠MCO = ∠COM
Since they are equal and angle M is 90°, then we can say they are both 45° and as such it is an isosceles triangle with two equal sides which are MO and MC. Thus;
MO = MC
Likewise doing the same for the other half and we have;
MO = DM.
Finally, we can say that MC = DM
Read more on cyclic quadrilaterals at; https://brainly.com/question/24368895
