circleseuclidean-geometrysolution-verificationtriangles
Question –
if C'AB , A'BC, B'CA are equilateral triangles drawn outwardly on sides of ABC. prove that centres X,Y,Z of equilateral triangles also form equilateral triangle…
My try –
Because all centres coincide in equilateral triangle so I try to prove that all angles are 60° of XYZ…and for that I make a lot of angle chasing, similarity,….etc but none of them works …
Any hints ?
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $\vert y-x\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, {x+y\over 2}, {x+y\over 2}$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$\left\vert {y\over 2}-{x\over 2}\right\vert={1\over 2}\vert y-x\vert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\vert y-x\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\over 2}$): if $r\in(-1,1)$ then for any $a$ we have
$$\lim_{n\rightarrow\infty}ar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
Since$$FH\parallel DB$$then$$\angle FHB=\angle DBJ=30^o$$And since$$HG\parallel CE$$then$$\angle BHG=\angle BCE=60^o$$
Therefore$$\angle FHG=30^o+60^o=90^o$$And since$$FH=\frac12 DB=\frac12 AB$$and$$GH=\frac12 EC=\frac12 BC$$then in area$$\triangle FHG=\frac14 \triangle ABC=8$$
Best Answer
The proof given below is done in two steps. In the first step, we prove that the circumcircles of the three equilateral triangles drawn outwardly on sides of an arbitrary $ABC$ have a common point. The second step consists of the arguments which prove that the centroids of the aforementioned equilateral triangles are the vertices of another equilateral triangle. In our proof, we use the fact that the centroid of an equilateral triangle coincides with the center of its circumcircle.
In $\mathrm{Fig. 1}$, we have drawn the circumcircles of the equilateral triangles $BPC$ and $CQA$. They intersect each other at $C$ and $F$ and, therefore, $CF$ is the common chord of the two shown circumcircles. To facilitates the proof, we need to join $BF$ and $AF$. Since $BPCF$ and $AFCQ$ are cyclic quadrilaterals, we have, $$\measuredangle CFB = \measuredangle AFC = 120^o.$$ Therefore, $\measuredangle BFA = 120^o$, which means that $ARBF$ is also a cyclic quadrilateral, because two of its opposite angles, namely $\measuredangle BFA$ and $\measuredangle ARB$, are supplementary. Consequently, the circumcircles of the three equilateral triangles drawn outwardly on sides of $ABC$ have $F$ as their common point.
$\mathrm{Fig. 2}$ shows the triangle obtained by joining the centroids $O_A$, $O_B$, and $O_C$. Our objective is to prove that each of the angles of this triangle is equal to $60^o$.
Now, draw the two common chords $BF$ and $CF$ of the circumcircle pairs {$O_A,\space O_C$} and {$O_A,\space O_B$} respectively. The sides of the triangle in question $O_A O_C$ and $O_A O_B$ are perpendicular to the two common chords $BF$ and $CF$ respectively. The lines $O_A O_B$ and $CF$ meets at $P$, while the lines $O_A O_C$ and $BF$ intersect each other at $Q$.
We have already shown that $\measuredangle CFB=120^o$. Now, consider the quadrilateral $O_A PFQ$. Its opposite angles $\measuredangle O_A PF$ and $\measuredangle FQO_A$ are supplementary. Therefore, $O_A PFQ$ is a cyclic quadrilateral. Consequently, its other pair of opposite angles $\measuredangle QO_A P$ and $\measuredangle PFQ$ are also supplementary. Therefore,
$$\measuredangle QO_A P + \measuredangle PFQ = 180^o,$$ which means, that $$\measuredangle QO_A P = 180^o - \measuredangle PFQ = 180^o - 120^o = 60^o.$$
Similarly, we can show that any of the other two angles of the triangle $O_A O_B O_C$ is equal to $60^o$, which is necessary and sufficient to conclude that it is indeed an equilateral triangle.