Question

The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

Answer 1

Consider ABCD as a cyclic quadrilateral with diagonals AC and BD intersecting at right angles

We know that OL ⊥ AB so that ∠O meets the line CD at the point M

From the figure we know that the angles in the same segment are equal

∠1 = ∠2

We know that ∠OLB = 90^o so we get

∠2 + ∠3 = 90^o …… (1)

We know that OLM is a straight line and ∠BOC = 90^o

So we get

∠3 + ∠4 = 90^o ……… (2)

It can be written as

∠2 + ∠3 = ∠3 + ∠4

On further calculation

∠2 = ∠4

So we get

∠1 = ∠2 and ∠2 = ∠4

It can be written as

∠1 = ∠4

We get

OM = CM

In the same way

OM = MD

So CM = MD

Therefore, it is proved that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.