Question

Prove that the perpendicular from the centre of a circle to a chord bisects the chord.

Answer 1

Hint: Here, we have to prove that the perpendicular from the centre of a circle to a chord bisects the chord. We have to prove that the perpendicular from the centre divided the chord into two equal parts. A chord of a circle is a straight line segment whose endpoints both lie on the circle.

Complete step by step solution:

C is a circle. Let O be the centre of the circle. Let OA and OB be the radius of the circle. AB is a chord of the circle such that \[OX \bot AB\]. OX is the perpendicular from the centre of the circle to the chord.
In \[\Delta OXA\] and \[\Delta OXB\],
OX is the perpendicular from the centre of the circle to the chord at \[90^\circ \],
\[\angle OXA = \angle OXB = 90^\circ \]
OX is the common side to both the triangles \[\Delta OXA\] and \[\Delta OXB\],
\[OX = OX\]
OA and OB be the radius of the circle, which is the hypotenuse of the triangles \[\Delta OXA\] and \[\Delta OXB\],
Radius of the circle is equal at all points. So,
\[OA = OB\]
So, by RHS rule, we will get
\[\Delta OAX \cong \Delta OBX\]
By Corresponding Parts of Congruent Triangles (CPCT) theorem, we have
\[AX = BX\]

Since AX is equal to BX, so the perpendicular from the centre of the circle bisects the chord at AB.

Therefore, we have proved that the perpendicular from the centre of a circle to a chord bisects the chord.

Note:
Here, we need to have knowledge about rules. If the hypotenuse and a side of a right- angled triangle is equivalent to the hypotenuse and a side of the second right-angled triangle, then the two right triangles are said to be congruent by RHS rule. CPCT theorem states that if two or more triangles which are congruent to each other are taken then the corresponding angles and the sides of the triangles are also congruent to each other.