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Home » AB and CD are two equal chords of a circle which is intersecting at P as shown in fig. P is joined to O, the center of the circle. So we need to Prove that OP bisects
AB and CD are two equal chords of a circle which is intersecting at P as shown in fig. P is joined to O, the center of the circle. So we need to Prove that OP bisects \angle CPB
Circles | Class 10 | Frank | Guides | ICSE | Maths
AB and CD are two equal chords of a circle which is intersecting at P as shown in fig. P is joined to O, the center of the circle. So we need to Prove that OP bisects \angle CPB

Construction: In the given figure draw perpendiculars OM and ON to AB and CD respectively.

Now, we have to consider the \vartriangle OMP and \vartriangle ONP,

OP = OP … [common side for both triangles of the given figure]

OM = ON … [distance of equal chords from the center of circle are equal]

\angle PMO=\angle PNO … [both the angles are equal that is {{90}^{\circ }}]

Therefore, we can say that \vartriangle OMP\cong \vartriangle ONP

So, \angle MPO=\angle NPO

Hence we have proved that, OP bisects \angle CPB.

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