Show that the points A, O, D are collinear iff triangle ABC is isosceles

Question

Let D be a point of the segment (BC). On the same side of the line BC we construct the isosceles triangles FDC and EBD, with FD = FC and EB = ED. We denote by O the middle of the segment EF and by A the intersection of the lines CF and BE. Show that the points A, O, D are collinear if and only if the triangle ABC is isosceles.

My idea:
I'm also going to put a photo below where I am going to show more exactly every idea I had. I went if ABC is isosceles I should demonstrate that A,O, D are collinear...I demonstrated it. (You can look in the picture below) Now I went backwards,if A,O,D. I dont know how I should demonstrate it. I thought of doing congruante triangles, but I dont find an angle or another line. I should demonstarte that AEDF is paralelogram.

Thank you so much! Hope one of you can help me! Any idea is welcome!

Answer 1

If the points $A$, $O$, $D$ are collinear:

Let $\angle BDE =\angle DBE = \alpha$ and $\angle CDF =\angle DCF = \beta$. Then $\angle EDF$ and $\angle EAF$ are both equal to $180^\circ - \alpha -\beta$. Now rotate triangle EDF through $180^\circ$ about $O$. Then $F$ maps to $E$, $E$ maps to $F$ and $D$ maps onto the line $OA$ - but because $\angle EDF=\angle EAF$ it must map onto $A$ itself. So $AEDF$ is a parallelogram and so $|AF|=|ED|=|EB|$ and $|AE|=|FD|=|FC|$. Thus $|AB| = |AE| +|EB| = |FC| + |AF| = |AC|$ as required.

If triangle $ABC$ is isosceles, then it is relatively easy to show that $A$, $O$, $D$ are collinear.

Answer 2

Here is a possible alternative (blind, analytic) solution, but consider accepting the solution of mcd first. We introduce algebra and work straightforward.

Let us denote by $E',O',A',F'$ the projections of the points $E,O,A,F$ on $BC$. Denote by $h_E$, $h_O$, $h_A$, $h_F$ the (lengths of the) segments $EE'$, $OO'$, $AA'$, $FF'$. Also let $x$ be $BE'=E'D$, and $y=DF'=F'C$, and $a= DA'$. We may and do assume that $x\le y$ (else reflect $D$ on $BC$ w.r.t. its mid point).

We consider the above situation with $a, h_A$ seen as unknowns. Then two equation in these two unknowns are immediately written: $$ \left\{ \begin{aligned} \frac {2x+a}{h_A} &=\frac{BA'}{AA'}=\frac{BE'}{EE'}=\frac x{h_E}\ ,\\ \frac {2y-a}{h_A} &=\frac{CA'}{AA'}=\frac{CF'}{FF'}=\frac y{h_F}\ , \end{aligned} \right. \qquad \text{ with solution } \qquad \left\{ \begin{aligned} a &= \frac{2xy(h_F-h_E)}{yh_E+xh_F}\ ,\\ h_A &= \frac{2h_Eh_F(x+y)}{yh_E+xh_F}\ . \end{aligned} \right. $$ We need only $$ \frac{a}{h_A}=\frac{xy/(x+y)}{h_Eh_F/(h_F-h_E)} \ . $$ Now the points $O,A,D$ are collinear, if( and only i)f we have the proportionality $DA':AA' =AO':OO'$. This means: $$ \frac{xy/(x+y)}{h_Eh_F/(h_F-h_E)} = \frac{(y-x)/2}{(h_E+h_F)/2} \qquad \text{ i.e. } \qquad \underbrace{\frac yx-\frac xy}_{(y^2-x^2)/(xy)} = \underbrace{\frac{h_F}{h_E} - \frac{h_E}{h_F}} _{(h_F^2-h_E^2)/(h_Eh_F)} \ . $$ Since the function $\displaystyle t\to t-\frac 1t$ is injective (strictly increasing), we obtain equivalently $y:x =h_F:h_E$.

So $O,A,D$ are collinear, iff $y:x =h_F:h_E$, iff $\Delta BEE'\sim\Delta CFF'$, iff $\hat B=\hat C$, iff $\Delta ABC$ is isosceles in $\bf A$.

$\square$