Q10 of 63 Page 145

If two equal chords of a circle intersect each other then prove that the ordered points of one chord are respectively equal to the corresponding points of the record chord.


Let AB and CD be two equal chords of a circle which are intersecting at a point E.


Construction:


Draw OF and OG perpendiculars on the chords


Join OE.


Consider ΔOFE and ΔOGE,


⇒ OF = OG [Equal chords]


⇒∠OFE = ∠OGE [Each 90°]


⇒ OE = OE [Common]


By RHS congruence rule,


⇒ ΔOFE ≅ ΔOGE


By CPCT,


⇒ FE = GE … (1)


Given AB = CD … (2)


⇒ 1/2 AB = 1/2 CD


⇒ AG = CF … (3)


Adding equations (1) and (3),


⇒ AG + GE = CF + FE


⇒ AE = CE … (4)


Subtracting equation (4) from (2),


⇒ AB – AE = CD - CE


⇒ BE = DE … (5)


From (4) and (5),


We can see that two parts of one chord are equal to two parts of another chord.


Hence proved


More from this chapter

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8

In a circle of radius 10 cm, the lengths of two parallel chords are 12 cm and 16 cm respectively. Find the distance between AB and CD. If chords (a) are on the same side of the centre (b) are on the opposite sides of the centre.

 9

The vertices of quadrilateral ABCD lie on a circle such that AB = CD. Then prove that AC = BD.

 11

Prove that the line segment joining the mid-points of two parallel chords of a circle passes through the centre of the circle.

 1

Write True/False for each and write the reason also for your answer.

(i) The angles subtended at any two points lying of the circle by every chord are equal.


(ii) In figure, AB is a diameter of a circle and C is any point on the circle. Then AC2 + BC2 = AB2.


(iii) In figure 12.48, if ∠ADE = 120°, then ∠EAB = 30°.


(iv) In figure 12.48, ∠CAD = ∠CED.