Question

The perpendicular from the centre of a circle to a chord bisects the chord​

Answer 1

Answer:

To prove that the perpendicular from the centre to a chord bisect the chord.

Consider a circle with centre at O and AB is a chord such that OX perpendicular to AB

To prove that AX=BX

In ΔOAX and ΔOBX

∠OXA=∠OXB [both are 90 ]

OA=OB (Both are radius of circle )

OX=OX (common side )

ΔOAX≅ΔOBX

AX=BX (by property of congruent triangles )

hence proved.

solution

Answer 2

Answer:

Theorem:

The perpendicular from the centre of a circle to a chord bisects the chord.

Proof:

Consider a circle with centre “O”.

AB is a chord such that the line OX is perpendicular to the chord AB.

(i.e) OX 1 AB

Now, we need to prove: AX = BX

To prove AX = BX, consider two triangles OAX and OBX.

OXA = OXB (<X = 90°)

OX = OX (Common side)

OA = OB (Radii)

By using the RHS rule, we can prove that the triangle OAX is congruent to OBX.

Therefore,

AOAX = AOBX

Hence, we can say that AX = BX (Using CPCT)

Thus, the perpendicular from the centre of a circle to a chord bisects the chord, is proved.

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