bisectioneuclidean-geometrygeometry
Chord $CD$ of a circle with $O$ as it center intersects diameter $AB$ at $I$ and $\angle AIC = 90°$. If chord $AE$ bisects $OC$ , then prove that $DE$ bisects $BC$.
My work:
Till now , I think this problem must have something to do with cyclic quadrilaterals or triangles. Maybe we need some similar of triangles to help.
I tried what I said , failed. Any hints please. Thank you very much!
Reflect $A$ across $D$ to a new point $Q$. Then $\triangle QDE\cong \triangle ECB$ (sas).
Since $$\angle QEC = \angle DEB $$ we see that quadrilateral $ACEQ$ is cyclic, so $$\angle QCE =\angle QAE = \angle DAE = \angle DNE$$ so $QC||DN$. Now since $D$ halves $AQ$ the line $DI$ is a middle line for triangle $QAC$ and thus it halves $AC$.
Let $DP$ intersect circle $I$ at $K$.
Notice that $RP\times PI =FP\times PE = KP\times PD$, therefore $R,K,I,D$ are co-cyclic. Also notive $IK=ID$, therefore all the red angles are equal immediately.
Now drop a perpendicular line from $A$ to BC, we have the two pink angles being equal as $ID$ is also perpendicular to $BC$.
Since the two cyan angles are equal and the ninty degree angles are equal, we have the two green angle at vertex $A$ being equal. Therefore the top red angle is equal to the top pink angle.
Look at the red and pink angles sharing edge $ID$, we know $PD$ is parallel to $AI$. Therefore $PD$ is perpendicular to $EF$.
Best Answer
A proof with analytic geometry. Set $O$ in the origin and assume, without loss of generality, that the circle has radius $1$. Align $OA$ on the $y$-axis. Then, $A$ is in $0,1$, $C$ is in $(-\sqrt{1-a^2}, a)$, and $D$ is in $(\sqrt{1-a^2},a)$.
The equation of the line containing the segment $AJ$ has intercept $1$ and is of the form $y=mx+1$. $J$ is in $(-\sqrt{1-a^2}/2, a/2)$. The slope $m$ can be determined by solving $a/2=-\sqrt{1-a^2}m/2+1$, which leads to $m= (a-2)(\sqrt{1-a^2})/(a^2-1)$. Solving a system with the line equation and the circle equation, we get the coordinates of $E$ are
$$\frac{2 (2 \sqrt{1 - a^2} -a \sqrt{1 - a^2})}{4a-5}, \frac{2 a^2-4a+3}{4a-5}$$
The equation of the line containing $ED$ is again obtained by another system built by accounting that the line passes through these two points. This leads to the equation
$$y=\frac{\sqrt{1 - a^2}}{3(1-a)} x +\frac{2a-1} {3}$$
Finally, the line containing the segment $BC$ has intercept equal to $-1$ and slope given by $-(a+1)/\sqrt{1-a^2}$. Solving a third system with the equations of the lines $ED$ and $BC$, we get that the coordinates of their intersection point $F$ are
$$F\left( -\frac{\sqrt{1-a^2}}{2}, \frac{a-1}{2}\right)$$
This completes the proof, because the coordinates of $F$ are midway with respect to those of $B$ and $C$.