Які центри трикутника збігаються в рівнобедреному трикутнику

A quick recap of the triangle centers:

$-$ circumcenter: the intersection of the perpendicular bisectors of the sides of the triangle (denoted $O$)

$-$ centroid: the intersection of the medians of the triangle (denoted $G$)

$-$ orthocenter: the intersection of the altitudes of the triangle (denoted $H$)

$-$ incenter: the intersection of the angle bisectors of the triangle (denoted $I$)

In the diagram, we can see that all the triangle centers appear to be collinear!

Let's think about why this might be true. They all appear to lie on the altitude from $B$ to $AC$, so it would be useful if we could show that for each one.

Having drawn in the altitude, we now see why the results should be true.

Since the triangle is isosceles, $\angle BAD=\angle BAC=\angle BCA=\angle BCD$. Finally, note that clearly $BD=BD$ and we have defined $D$ so that $\angle BDA=\angle BDC=90^\circ$. But then by AAS we must have that $\triangle ABD\cong \triangle CBD$!

This tells us many things: $-$ We must have $AD=CD$ then, so $D$ is the midpoint of $AC$ $-$ We must have $\angle DBA=\angle DBC$, so $BD$ bisects $\angle ABC$ $-$ The altitude from $B$ must also be the perpendicular bisector of $AC$, since it goes through the midpoint of $AC$ (namely, $D$) and is perpendicular to $AC$

So then this altitude turns out to be not just an altitude, but also an angle bisector, a perpendicular bisector, and a median!

From this, we conclude that the triangle centers must all be collinear in an isosceles triangle.

One final note: what if we have an equilateral triangle? This is essentially a more powerful version of the isosceles triangle, in that all pairs of sides are equal. It turns out that in an equilateral triangle, we actually have that all triangle centers are in the same place (think about why this is true).