**General enunciation**: We have to show that of the two chords of a circle, the greater chord is nearer to the centre than the shorter.

**Particular enunciation**: Let, O is the centre of the circle ABCD. AB and CD are two chords where AB > CD. OE and OF are the perpendiculars from centre O to the chords AB and CD respectively. Let us show that, OE < OF.

**Construction**: (O, A) and (O, C) are joined.

**Proof**: Because O is the centre of the circle and OE ⊥ AB then we get AE = ½ AB.

Similarly, O is the centre of the circle and OF ⊥ CD then we get CF = ½ CD.

Now, in right angle triangle **∆ **OAE, OA is the hypotenuse.

∴ OA^{2} = OE^{2} + AE^{2} ………(1)

Again, in right angle triangle **∆ **OFC, OC is the hypotenuse.

∴ OC^{2} = OF^{2} + CF^{2} ………(2)

But, OA and OC are the radii of the same circle.

∴ OA = OC

So, from equation (1) and (2) we get,

OE^{2} + AE^{2} = OF^{2} + CF^{2}

⟹ AE^{2} – CF^{2} = OF^{2} – OE^{2} ……… (3)

But, according to the given question, AB > CD

⟹ ½ AB > ½ CD

⟹ AE > CF

⟹ AE^{2} >CF^{2}

⟹AE^{2} – CF^{2} > 0

⟹ OF^{2} – OE^{2} > 0 [∵AE^{2} – CF^{2} = OF^{2} – OE^{2}]

⟹ OF > OE

That is, OE < OF. **(Proved)**

**Problem**: A circle passes through the vertices of a right angled triangle. Show that, the centre of the circle is the middle point of the hypotenuse.

**General enunciation**: A circle passes through the vertices of a right angled triangle. We have to show that, the centre of the circle is the middle point of the hypotenuse.

**Particular enunciation**: Let O is the centre of the circle ABC, which passes through three vertices A, B and C of the right angle triangle **∆ **ABC. Let us show that the centre of the circle O is the middle point of the hypotenuse AB.

**Construction**: (O, C) are joined.

**Proof**: In **∆ **BOC, OB = OC [radii of the same circle]

∴ ∠OCB = ∠OBC

⟹∠ACO = ∠ABC

Again, In **∆ **AOC, OA = OC [radii of the same circle]

∴ ∠OCA = ∠CAO

⟹∠BCO = ∠BAC

In **∆ **ABC, ∠ACB + ∠ABC + ∠BAC = 180^{0}

⟹ ∠ACB + ∠ACO + ∠BCO = 180^{0} [∵∠ACO = ∠ABC, BCO = ∠BAC]

⟹ ∠ACB + ∠ACB = 180^{0}

⟹ 2 ∠ACB = 180^{0}

∴ ∠ACB = 90^{0}

Since ∠ACB = 180^{0} then AB is the hypotenuse of the **∆ **ABC.

Now, AB = OA + OB

⟹ AB = OA + OA [OA and OB are radii of the same circle]

⟹ 2OA = AB

∴ OA = ½ AB

That is, O is the middle point of AB.** (Showed).**