Second Balkan Mathematical Olympiad

[nafistiham]

let $O$ be the centre of the circle through the points $A,B,C$.Let $D$be the mid point of $AB$.Let $E$ be the centroid of the triangle $ACD$.Prove that, the line $CD$ will be perpendicular to the line $OE$ iff $AB=AC$

the book provides the solution in vector could someone do it in the geometric way?

$7777^{th}$

[Avik Roy]

প্রশ্নই তো অসম্পূর্ণ মনে হচ্ছে!!!

[nafistiham]

Prove that, the line $CD$ will be perpendicular to the line $OE$ iff $AB+AC$

sorry.
it should have been like this.

Prove that, the line $CD$ will be perpendicular to the line $OE$ iff $AB=AC$

[Phlembac Adib Hasan]

Geo 3.pdf

(118.17 KiB) Downloaded 189 times

It's very easy to prove $ OE\perp CD $ when $AB=AC$.So I'm giving its inverse case.
My Proof :

Given that $ OE\perp CD $.We have to prove $AB=AC$.Let $AB\cap CE=F$.(Not shown in the figure).
From Stewart's theorem,\[AF.BC^2+BF.AC^2=AB(CF^2+AF.FB)\]
\[\Rightarrow AB(\frac{BC^2}{4}+\frac{3AC^2}{4})=AB(CF^2+\frac{3AB^2}{16})\]
\[\Leftrightarrow (\frac{BC^2}{4}+\frac{3AC^2}{4})=(CF^2+\frac{3AB^2}{16})\]
\[\Leftrightarrow CF^2=\frac{BC^2}{4}+\frac{3AC^2}{4}-\frac{3AB^2}{16}\]
\[\Leftrightarrow \frac{4CF^2}{9}=\frac{BC^2}{9}+\frac{AC^2}{3}-\frac{AB^2}{12}\]
\[\Rightarrow CE^2=\frac{BC^2}{9}+\frac{AC^2}{3}-\frac{AB^2}{12}\]
\[\Rightarrow CE^2-DE^2=\frac{BC^2}{9}+\frac{AC^2}{3}-\frac{AB^2}{12}-\frac{BC^2}{9}\]
\[\Rightarrow CE^2-DE^2=\frac{AC^2}{3}-\frac{AB^2}{12}\]
Notice that \[OD^2=AO^2-\frac{AB^2}{4}=R^2-\frac{AB^2}{4}\]
\[\Rightarrow CO^2-OD^2=\frac{AB^2}{4}\]
As $ OE\perp CD $, from perpendicular lemma,\[CO^2-OD^2=CE^2-DE^2\]
\[\Rightarrow \frac{AB^2}{4}=\frac{AC^2}{3}-\frac{AB^2}{12}\]
\[\Rightarrow \frac{AB^2}{3}=\frac{AC^2}{3}\]
\[\Rightarrow AB=AC\]

[nafistiham]

thanks for the nice proof. (and i am uploading the figure as a jpeg file, so that none has to download it. )

from a previous post by phlembac adib hasan

Geo 3.JPG (40.55 KiB) Viewed 1835 times

$400^{th}$