Problem 8:
(a) Prove that in a circle, the line connecting the center and the midpoint of a chord intersects the chord at right angles.
(b) Suppose the points $A, B, C, D$ and $E$ are chosen counter-clockwise on a circle. $X$ is a point inside the circle, and $AX=CX=EX$. Prove that $BX=DX$.
BdMO National Junior 2011/8
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Re: BdMO National Junior 2011/8
(a)solution:suppose,'$c$' is a circle.$O$ is it's centre.Draw a chord $AB$. $D$ is the midpoint of $AB$.Join O,$D.OA$=$OB$,$AD=BD$ and $OD$ is the common side of $\triangle OAD$ and $\triangle ODB$ .So $\triangle OAD \cong \triangle ODB$. So $\angle ADO=\angle BDO=90^{\circ}$
(b)solution: $X$ is the centre of the circle.So $BX=DX$
(b)solution: $X$ is the centre of the circle.So $BX=DX$
Last edited by *Mahi* on Wed Jan 29, 2014 8:01 pm, edited 1 time in total.
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Re: BdMO National Junior 2011/8
Is the solution of (b) correct!
"Questions we can't answer are far better than answers we can't question"
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Re: BdMO National Junior 2011/8
Yes. $X$ lies on the perpendicualr bisector of $A$C and $AE$.So $X$ is the center of the circle.tanmoy wrote:Is the solution of (b) correct!