Question

Prove that all the chords of a circle through a given point within it l, the least is one which is bisected at that point.

Answer 1

let the center oof the circle be oand the radius be r
let n be a point on a circle through witch cord ab passes and n is the mid point of ab
let another chord cdalso passes through n.
we need to prove that cd > ab
draw om perpenducular to cd and join on.
in right triangle omn,we have
ON > OM since ON is the hypotenus
thus chord cd is closer to o as comper to chord ab
 Thus, CD > AB ..... proved

Answer 2

hey!

we need to proove chord AB is smaller than chord CD bcoz AB is bisected by point M not CD is bisected we can clearly see in the figure

construction : Join ON and OM

to proove : AB<CD

in a right angled ∆ a hypotensue is always larger

so in right ∆ONM

OM is the hypotensue so OM>ON

here ON is smaller so by thorem we know that of any two chords of a circle the one which is nearer to the centre is longer

so CD>AB

hence prooved !