A nice property of a triangle With side 1,2,2

Question

About a year ago, while using GeoGebra, I discovered a beautiful property of a triangle with sides 1,2,2.

It is that many of the important centers of this triangle are located at equal distances in a row, so much so that I called this triangle the “numbers line triangle.”

You can consider the following two images:

$G$ is the point of intersection of the heights of $∆XYZ$

$C$ is the point of intersection of the bisectors of $∆X^{'}Y^{'}Z^{'}$

$F$ is the intersection of the bisectors of $∆XYZ$

$B$ is the center of the circle $XMPDQN$

$E$ is the point of convergence of the averages of $∆XYZ$

$A$ is the point of convergence of the heights of $∆X^{'}Y^{'}Z^{'}$

$D$ is the point of convergence of the $∆XYZ$ axes

$XA=AB=BC=CD=DE=EF=FG$

My question consists of two parts:

Is this feature already discovered?

How do we prove that anyway?

Answer 1

Let us introduce all the points in a clear order, and show the claimed properties one by one. To have usual, standard notations, i will switch in notation from $\Delta XYZ$ and $\Delta X'Y'Z'$ to $\Delta ABC$ and $\Delta A'B'C'$. Then instead of the points $A,B,C,D,E,F,G$ i will use (in order to avoid collisions) $P_A,P_B,P_C,P_D,P_E,P_F,P_G$ and also their standard ETC notations. To avoid fractions i will work with the $\color{red}{\bf 2-4-4}$ triangle!

So we want to show:

Proposition: Let $\Delta ABC$ be a triangle with sides $BC=a=2$, $AC=b=4$, $AB=c=4$. Let $(O)$ be the circumcircle of $\Delta ABC$, its center is $O$. Let $I,H,G$ be respectively its incenter, orthocenter, and centroid. Let $I',H',G'=G$ be respectively incenter, orthocenter, and centroid in $\Delta A'B'C'$.

Let $A'$ be so that $A'B\| AC$ and $A'C\|AB$. Let $B'$ be so that $B'A\| BC$ and $B'C\|AB$. Let $C'$ be so that $C'A\| BC$ and $C'B\|AC$.

Let $BB^\maltese$, $CC^\maltese$ be the heights from $B,C$ in $\Delta ABC$. They intersect in $H$. Let $B'B''$, $C'C''$ be the heights from $B',C'$ in $\Delta A'B'C'$. They intersect in $H'$.

Let $B_G,C_G$ be the mid points of $AC$, $AB$. Let $BIB_I$, $CIC_I$ be the angle bisector cevians in $\Delta ABC$.

Let $M,M'$ be the incenter, respectively the centroid in $\Delta ACB'$. Reflected w.r.t. $AT$ are the points $N,N'$, corrsponding centers for $\Delta ABC'$.

We define the points $P_A,P_B,P_C,P_D,P_E,P_F,P_G,P_H$ as in the picture to be respectively $H'$, mid point of $PO$, $I',O,G,I,H$, and the point on the circumcircle diametrically opposed to $A$.

Then:

• $(i)$ The points $B'',C''$ are on the circumcircle $(O)=(ABC)$.
• $(ii)$ The angle bisectors $B'I'$ and $C'I'$ pass through $C_G$, respectively $B_G$.
• $(iii)$ $M$ is on $C'C''$, $N$ is on $B'B''$. The points $M,M';N,N';O$ are on the circumcircle $(P_B)=(AB_GC_G)$.
• $(iv)$ We have the equality of (measures of) segments: $$AP_A=P_AP_B=P_BP_C=P_CP_D=P_DP_E=P_EP_F=P_FP_G=P_GP_H\ .$$

Proof: Let $T$ be the mid point of $BC$. We have two known sides in $\Delta ABT$, so the third one is $AT=\sqrt{4^2-1^2}=\sqrt{15}$. The following triangles are by construction similar, $\Delta ABT\sim \Delta B'C'B''\sim B'H'A$, so we obtain $AH':2=AH':AB'=BT:AT=1:\sqrt{15}$, so $$ AH'=\frac 2{\sqrt {15}}\ .$$ The triangle $\Delta B''B'C'$ has a right angle in $B''$, and $A$ is the mid point of the hypotenuse, so $AB''=AB'=BC$. Since $B''B\|AC$ the trapezium $BB''AC$ is isosceles, thus cyclic, so $B''$ is on the circumcircle $(O)$. This shows $(i)$.

Let $J$ be the mid point of $B'C$. Then $BCJC_G$ and $C_GJB'A$ are congruent rhombi, since all sides have length $2$, and opposite sides are respectively parallel. In particular $B'C_G$ is the angle bisector in $B'$ and in $C_G$ in the last rhombus. So $B'I'$ passes through $C_G$. This shows $(ii)$. Note also that in the first rhombus the diagonals are perpendicular, so the median $CGC_G$ is perpendicular on the angle bisector $BIB_I$. We record the fact, that $P_C=I'$ is the incenter also in $\Delta AC_GB_G$, a triangle similar to $\Delta ABC$.

The right triangles $\Delta C'B'C''$ and $\Delta ACT$ are congruent, same angles and an hypotenuse of size $4$ each. So $B'C''=CT=1$, showing that $C''$ is the mid point of $B'J$. In particular we have the congruent triangles $\Delta C'JB'\equiv \Delta ABC$.

Before we show $(iii)$, let us cover the location of some points in $(iv)$. For each point $P_?$ it is convenient to compute the distance $AP_?$ as a multiple of the height $AT$. We have relatively easy:

• For the incenter $I=P_F$ using the fact that $BI$ is angle bisector in $\Delta ABT$ we have $AI:IT=BA:BT=4:1$, so $$ \frac{AP_F}{AT}=\frac 45=\frac{\bbox[yellow]{12}}{15}\ . $$
• For the centroid $G=P_E$ we have directly $$ \frac{AP_E}{AT}=\frac 23=\frac{\bbox[yellow]{10}}{15}\ . $$
• For the circumcenter $O=P_D$ we compute the circumradius. Either solving an equation in the unknown $R=AO=OB=OC$ from Pythagoras in $\Delta BOT$, $R^2=1^2+(\sqrt {15}-R)^2$, or by computing the area of the triangle as $S=AT\cdot BT=\sqrt {15}$, and using $AO=R=(abc)/(4S)=8/sqrt{15}=\frac8{15}\cdot AT$, so $$ \frac{AP_D}{AT}=\frac{\bbox[yellow]{8}}{15}\ . $$
• $P_B$ is the mid point of $AO$, and $O$ the mid point of $AP_H$so we have immediately: $$ \begin{aligned} \frac{AP_B}{AT} &= \frac {\bbox[yellow]{4}}{15}\ ,\\ \frac{AP_H}{AT} &= \frac {\bbox[yellow]{16}}{15}\ . \end{aligned} $$
• We have seen in $(ii)$ that $P_C=I'$ is the incenter of $\Delta AC_GC_B$, which is similar to $\Delta ABC$. We know the location of $I$ in the last triangle w.r.t. the height $AT$, so from $AI'=\frac 12AI$ we obtain $$ \frac{AP_C}{AT}=\frac{\bbox[yellow]{6}}{15}\ . $$
• The point $P_G=H$ is a vertex in the triangle $\Delta CTH$, and $\Delta CTH\sim \Delta CC^\maltese B\sim \Delta ATB$, so $HT:1=HT:CT=BT:AT=1:AT=AT:AT^2=AT:15$. This leads to $HT:AT=1:15$, so $$ \frac{AP_G}{AT}=\frac{\bbox[yellow]{14}}{15}\ . $$ The triangle $\Delta A'B'C'$ is similar to $\Delta ABC$, obtained by a factor-two-dilation. So $H'A=2HT$, which leads to $$ \frac{AP_A}{AT}=\frac{\bbox[yellow]{2}}{15}\ . $$ This concludes $(iv)$.

It remains $(iii)$ now. By definition the points $M,M';N,N'$ are on the symmetry axes of $\Delta ACB'$, $\Delta ABC'$, and are mirrored in $AT$, so $MNN'M'$ is a rectangle. As seen in a note after $(ii)$ there is a congruence of triangles
$\Delta C'JB'\equiv \Delta ABC\equiv CAB'$. The positions of the first and last triangle are reflected w.r.t. to the angle bisector $B'MI'$, because the common angle in $B'$ switches the sides and $C,C'$, respectively $A,J$ exchange places. Since $M$ is on the $C$-height (and on the reflection line) in the one triangle, it is also on the reflected height, which is $C'C''$. Similarly $N\in B'B''$.

Now consider the points $M,M'$ in $\Delta CAB'$. They correspond in the congruent triangle $\Delta ABC$ to the points $I,G$, in particular $MM'=IG=AP_A=P_AP_B=P_BP_C=P_CP_D$. These segments also have the same direction (are parallel). Observe now that $P_B$ projects on the segment $MM'$ in its mid point. This is so because the distances from $P_B,M,M'$ to the line $C'AB'$ are known, they are $AT$ times respectively $4/15$, $3/15$ (as $I$), $5/15$ (as $GT$). We can even compute the distance $P_BM$, its projection on $MM'$ is $MM'/2=P_AP_B/2=AT/15=1/\sqrt{15}$, its projection on $AB'$ is $AB'/2=1$, so $$ P_BM^2=\frac 1{15}+1=\frac{16}{15}=P_BA^2\ . $$ Putting all toghether we obtain $(iii)$.

$\square$

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Answer 2

This is a brief proof, I may add details later

Let's consider standard notation, then

We just need to prove that $GI=HI$ because $OH=3OG$

Applying the angular conjugation property and the inner bisector theorem, we get this

The rest of the centers are easy to show similarly because the other triangle is an enlargement of the same triangle