Question

Prove that the diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

Answer 1

Hint: We will prove this by the congruence of two triangles. By fulfilling the side-side-side (SSS) condition of congruence we can conclude the required result.

Condition of congruence:

(i)SSS (side-side-side)

(ii)SAS (side-angle-side)

(iii)ASA (angle-side-angle)

(iv)AAS (angle-angle-side)

(v)RHS (right angle -Hypotenuse-side)

Complete step by step answer:

We need to prove the given statement for doing that we need to draw the circle.

Here we draw the circle with the diameter AB which bisects the chord MN at P. O is the centre of the circle.

We need to prove that AB bisects \[\angle MON\].

Now we have, OM and ON are the radius of the circle so these two are equal.

 Therefore, we get, In \[\Delta MOP\] and \[\Delta NOP\]

OM=ON

OP is the common side (So, OP=OP).

MP=PN (given) [since AB bisects MN at P]

Hence by SSS condition the triangles \[\Delta MOP\]and \[\Delta NOP\]are congruent.

i.e. \[\Delta MOP \cong \Delta NOP\]

Hence, \[\angle MOP = \angle NOP\]

i.e. the angles are equal.

Therefore, we get, AB bisects\[\angle MON\].

Hence, the diameter of a circle that bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle. (Proved)

Note: Congruence of triangles:

Two triangles are said to be congruent once we get, all three corresponding sides are equal and all the three corresponding angles are equal in the measure. These triangles can be slides, rotated, flipped, turned to look identical. If repositioned, they coincide with each other. The symbol of congruence is \[ \cong \].