Q8 of 31 Page 65

I prove that, the two intersecting chords of any circle cannot bisect each other unless both of them are diameters of the circle.


Given, AB and CD are Diameters.


To prove: OA = OB, OC = OD


Construction: Point D joined with B, Point A joined with C.


DAC = 900 (Rectangle)


ACB = 900 (Rectangle)


CBD = 900 (Rectangle)


BDA = 900 (Rectangle)


ACBD is Rectangle, So AD = CB, AD BCand BD = AC,BD AC


In OCB and ODA


OCB = ODA (Interior angles, BC AD)


BC = AD (Rectangle)


OBC = OAD (Interior angles, BC AD)


BCA SA Congruency


In OCB ODA


Hence Using CPCT, OA = OB, OD = OC


Hence Proved.


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