Problem 3. Let ABC be a right-angled triangle with ∡BAC=90°, let D be the foot of the altitude from A to BC, and let E be the midpoint of DC. The circumcircle of ΔABD intersects AE again at point F. Let X be the intersection of the lines AB and DF. Prove that XD ¯ = XC ¯ .

Solution 1Solution 2

Solution 1

We must prove that $XE\perp CD$ i.e, $AD\parallel XE$. So let $K$ be the midpoint of $AC$, then we will prove that the lines $KE$, $FD$ and $AB$ are concurrent. Firstly $\angle CKE=\angle CAD=\angle CBA$, so quadrilateral $KEBA$ cyclic.
Since $K$ midpoint, $KA=KD=KC$. Then $KD$ is also tangent to $(ABDF)$, because $KA=KD$ and $KA$ tangent to it. So $\angle KAF=\angle DAF=\angle FEK$, i.e quadrilateral $DFEK$ is cyclic. Lines $KE$, $FD$ and $AB$ are radical axis of circles $(EKAB)$, $(FABD)$ and $(KEFD)$, so they are concurrent. So we are done!

Solution 2

Here is possibly the most straightforward solution to this problem. The desired reduces to showing that $XE\parallel AD$, or $\frac{AF}{EF}=\frac{DF}{FX}$. We compute each of these ratios by hand.

Let $AB=x$ and $AC=y$. Then $BC=\sqrt{x^2+y^2}$ and


$$BD=\frac{x^2}{\sqrt{x^2+y^2}}, \quad DE=EC=\frac{y^2}{2\sqrt{x^2+y^2}}, \quad AD=\frac{xy}{\sqrt{x^2+y^2}},$$

so
$$AE=\sqrt{DE^2+AD^2}=\frac{y\sqrt{4x^2+y^2}}{2\sqrt{x^2+y^2}}.$$

Now by power of a point


$$ED\cdot EB=EF\cdot EA\iff\frac{y^2\left(2x^2+y^2\right)}{4\left(x^2+y^2\right)}=\frac{y\sqrt{4x^2+y^2}}{2\sqrt{x^2+y^2}}\cdot EF\iff EF=\frac{y\left(2x^2+y^2\right)}{2\sqrt{\left(x^2+y^2\right)\left(4x^2+y^2\right)}}.$$

Therefore


$$AF=AE-EF=\frac{x^2y}{\sqrt{\left(x^2+y^2\right)\left(4x^2+y^2\right)}}$$

Now by Menelaus in $\triangle BEA$ we find that


$$\frac{BD}{ED}\cdot\frac{EF}{AF}\cdot\frac{AX}{BX}=1\iff\frac{2x^2}{y^2}\cdot\frac{2x^2+y^2}{2x^2}\cdot\frac{AX}{BX}=1,$$

whence $\displaystyle \frac{BX}{AX}=\frac{2x^2+y^2}{y^2}$ and $AX=\frac{y^2}{2x}$.

Finally, by Menelaus in $\triangle BDX$ it follows that


$$\frac{BE}{DE}\cdot\frac{DF}{XF}\cdot\frac{XA}{BA}=1\iff\frac{2x^2+y^2}{y^2}\cdot\frac{DF}{FX}\cdot\frac{y^2}{2x^2}=1\iff\frac{DF}{FX}=\frac{2x^2}{2x^2+y^2}.$$


Due to the lengths computed before, the above can easily be equated to $AF/EF$, as desired.

Solution 3

Position the triangle such that $BC$ is horizontal, and $D$ is the origin. As $BAC$ is a right triangle, we can let

 


$$B=(-1,0),A=(0,r),C=(r^2,0).$$

We then get


$$E=\left(\frac{r^2}2,0\right)$$

and the circumcircle of $ABD$ is centred at the midpoint of $AB$, which is


$$\left(-\frac12,\frac r2\right).$$

As this passes through $(0,0)$, the equation of the circle is


$$x^2+x+y^2-ry=0$$

We have line $AE$ passes through $(0,r)$ and $\left(\frac{r^2}2,0\right)$, so its equation is


$$rx+\frac{r^2}2y=\frac{r^3}2$$

This gives


$$x=\frac{r^2}2-\frac r2y$$

so then the intersections of this with the circle becomes


$$\left(\frac{r^2}2-\frac r2y\right)^2+\frac{r^2}2-\frac r2y+y^2-ry=0$$

$$\frac{r^2}4y^2-\frac{r^3}2y+\frac{r^4}4+\frac{r^2}2-\frac r2y+y^2-ry=0$$

$$(\frac{r^2}4+1)y^2+(-\frac{r^3}2-\frac{3r}2)y+\frac{r^4}4+\frac{r^2}2=0$$

Multiply by $4$ to get


$$(r^2+4)y^2+(-2r^3-6r)y+r^4+2r^2=0$$

Factor out $y-r$ to get


$$(y-r)((r^2+4)y-r^3-2r)$$

As $F$ is not $A$, this means that the $y$-coordinate of $F$ is $\frac{r^3+2r}{r^2+4}$. As $x=\frac{r^2}2-\frac r2y$, we have


$$x=\frac{r^2}2-\frac r2\frac{r^3+2r}{r^2+4}=\frac{r^4+4r^2}{2r^2+8}-\frac{r^4+2r^2}{2r^2+8}=\frac{r^2}{r^2+4}$$

Then, the line $DF$ has the equation


$$\frac yx=\frac{\frac{r^3+2r}{r^2+4}}{\frac{r^2}{r^2+4}}=\frac{r^2+2}r$$

The line $AB$ has the equation $y=rx+r$, so we have


$$\frac{r^2+2}rx=rx+r$$

$$(r^2+2)x=r^2x+r^2$$

$$2x=r^2$$

$$x=\frac{r^2}2$$

Notice that the perpendicular bisector of $DC$, which are $(0,0)$ and $(r^2,0)$, is obviously just $x=\frac{r^2}2$, so we see that the intersection of $DF$ and $AB$ lies on the perpendicular bisector of $DC$, so $XD=XC$.