Question

Two equal chords AB and CD of a circle with centre O intersect each other at right angle at point P. If OM ⊥ AB and ON ⊥ CD, prove that OMPN is a square.

Answer 1

Given:

AB and CD are equal chords of a circle with centre O.

AB and CD intersect at right angle at P, so AB ⟂ CD at P.

OM ⟂ AB   ...(M is foot of perpendicular from O to AB)

And ON ⟂ CD   ...(N is foot of perpendicular from O to CD)

To Prove: OMPN is a square.

Proof (Step-wise):

1. Since OM ⟂ AB and ON ⟂ CD, M lies on AB and N lies on CD by definition.

2. Equal chords are equidistant from the centre, hence OM = ON.

3. OM ⟂ AB and CD ⟂ AB.   ...(Given AB ⟂ CD) 

So, OM and CD are both perpendicular to AB; therefore, OM || CD. 

In particular, OM || PN.   ...(∵ PN is a part of CD)

Similarly, ON || AB, so ON || MP.   ...(∵ MP is a part of AB)

Thus, opposite sides OM and PN are parallel, and opposite sides ON and MP are parallel.

4. Consider right triangles ΔOMP and ΔONP:

∠OMP = 90° and ∠ONP = 90°   ...(By construction)

OM = ON   ...(From step 2)

OP is common.

Hence, ΔOMP ≅ ΔONP by RHS congruence, so MP = PN and ∠OPM = ∠OPN.

5. Because MP ⟂ PN (AB ⟂ CD at P), the angle MPN = 90°.

From congruence (step 4), we get

∠OPM = ∠OPN = 45°

Thus, ΔOMP is a right triangle with one acute angle of 45°, so it is a 45° – 45° – 90° triangle. 

Therefore, OM = MP.

6. From steps 4 and 5,

We now have OM = MP = PN = NO   ...(OM = ON from step 2, MP = PN from step 4, OM = MP from step 5).

7. We already showed opposite sides are parallel (step 3) and adjacent sides are perpendicular OM ⟂ MP and MP ⟂ PN etc. 

Therefore, OMPN has four equal sides and all angles are 90°, i.e., OMPN is a square.