Question

In fig. 15.101, AB and CD are two chords of a circle, intersecting each other at P such that AP = CP. Show that AB = CD.

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Answer 1

Given:-

→AB and CD are two chords of a circle

→AB and CD intersect each other at P.

→AP = CP

To show:-

→AB = CD

Solution:-

It is given that chords AB and CD are intersecting each other inside the circle at P.

Thus :-

=> AP×PB = CP×PD

=> AP/CP = PB/PD

∵ AP = CP [Given] ----(1)

∴ PB = PD ----(2)

Now, on adding eq.1 and eq.2, we get:-

=> AP + PB = CP + PD

=> AB = CD [∵AP+PB = AB

CP+PD= CD]

Hence,proved !!!

Some Extra Information:-

•Line segment joining the centre to any point on a circle, is a radius of the circle.

•A chord of a circle is a straight line segments whose endpoints lie in the circle.

•A chord of a circle, which is twice as the radius, is the diameter of the circle.

•Segment of a circle is a region between an arc and chord of the circle.

Answer 2

Given:-

→AB and CD are two chords of a circle

→AB and CD intersect each other at P.

→AP = CP

To show:-

→AB = CD

Solution:-

It is given that chords AB and CD are intersecting each other inside the circle at P.

Thus :-

=> AP×PB = CP×PD

=> AP/CP = PB/PD

∵ AP = CP [Given] ----(1)

∴ PB = PD ----(2)

Now, on adding eq.1 and eq.2, we get:-

=> AP + PB = CP + PD

=> AB = CD [∵AP+PB = AB

CP+PD= CD]

Hence,proved !!!

Some Extra Information:-

•Line segment joining the centre to any point on a circle, is a radius of the circle.

•A chord of a circle is a straight line segments whose endpoints lie in the circle.

•A chord of a circle, which is twice as the radius, is the diameter of the circle.

•Segment of a circle is a region between an arc and chord of the circle.