prove that if any straight line passing through the centre of a circle bisects any chords which is not a diameter. then the straight line will be perpendicular on that chord
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Let R be the centre of the circle.
Let AB be a straight line passing through R and bisecting a chord CD
so that BC=BD
Join RC and RD
In triangle RBC and triangle RBD :
RB=RB (common)
BC=BD(given)
RC=RD(radius of a circle)
By SSS(side side side) congruency, RBC is congruent to RBD
By Cpct, angle RBC = angle RBD
CD is a straight chord
RBC+RBD=180(Linear pair)
2RBC=180 ( since RBC=RBD)
2RBD=180
RBC=RBD=90
AB is a straight line
ABC=RBC=90
so , AB is perpendicular to CD
Hence,
if any straight line passing through the centre of a circle bisects any chords which is not a diameter, then the straight line will be perpendicular on that chord.
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