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Home / Geometry / Show that of the two chords of a circle, the greater chord is nearer to the centre than the shorter.

Show that of the two chords of a circle, the greater chord is nearer to the centre than the shorter.

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.

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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.

∴ OA2 = OE2 + AE2 ………(1)

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

∴ OC2 = OF2 + CF2 ………(2)

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

∴ OA = OC

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

OE2 + AE2 = OF2 + CF2

⟹ AE2 – CF2 = OF2 – OE2 ……… (3)

But, according to the given question, AB > CD

⟹ ½ AB > ½ CD

⟹ AE > CF

⟹ AE2 >CF2

⟹AE2 – CF2 > 0

⟹ OF2 – OE2 > 0   [∵AE2 – CF2 = OF2 – OE2]

⟹ 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.

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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 = 1800

⟹ ∠ACB + ∠ACO + ∠BCO = 1800 [∵∠ACO = ∠ABC, BCO = ∠BAC]

⟹ ∠ACB + ∠ACB = 1800

⟹ 2 ∠ACB = 1800

∴ ∠ACB = 900

Since ∠ACB = 1800 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).

 

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