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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.

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.

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).

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...