Angle Chasing Basics

Through vertex $A$ draw a line parallel to $BC$. Since $AB$ is a transversal of the parallel lines, the angle $\beta$ at $A$ and the angle $\beta$ at $B$ are alternate, hence equal. Similarly, via the transversal $AC$, the angles $\gamma$ at $A$ and at $C$ are alternate. The angles $\beta$, $\alpha$, $\gamma$ lie next to each other along the line through $A$, so $\alpha + \beta + \gamma = 180^\circ.$

Proof 1 (mathematical induction). Base case $n=3$: the sum of the angles of a triangle is $180^\circ = (3-2)\cdot 180^\circ$. Inductive step: let $A_1 A_2 \ldots A_{n+1}$ be a convex $(n+1)$-gon. The diagonal $A_1 A_n$ splits it into a triangle $A_1 A_n A_{n+1}$ and a convex $n$-gon $A_1 A_2 \ldots A_n$: the angles at vertices $A_2, \ldots, A_{n-1}$ belong entirely to the $n$-gon, the angle at $A_{n+1}$ belongs entirely to the triangle, and the angles at $A_1$, $A_n$ are split between the two figures so that their parts add up to the original interior angles of the $(n+1)$-gon. The angle sum of the $(n+1)$-gon is therefore

Proof 2 (fan triangulation). From vertex $A_1$ draw diagonals to every non-adjacent vertex $A_3, A_4, \ldots, A_{n-1}$. This produces $n-2$ triangles that cover the interior of the $n$-gon without overlap. At each vertex $A_k$, the angles of the adjacent triangles at $A_k$ assemble exactly into the interior angle of the $n$-gon at $A_k$, so the sum of the angles of all triangles equals the sum of the interior angles of the $n$-gon, i.e. $(n-2)\cdot 180^\circ$.

Poznámka. Proof 2 is an unrolled version of Proof 1: induction cuts off triangles one by one, and here we see them all at once.

Proof 3 (interior point). Choose a point $O$ in the interior of the $n$-gon and connect it to every vertex. This produces $n$ triangles with a total angle sum of $n\cdot 180^\circ$. At each vertex $A_k$ of the $n$-gon, the triangle angles at $A_k$ together form the entire interior angle of the $n$-gon there; the triangle angles at $O$ together form a full angle of $360^\circ$. So the sum of the interior angles of the $n$-gon is

Poznámka (non-convex polygons). The formula $(n-2)\cdot 180^\circ$ also holds for non-convex polygons, provided we measure all interior angles toward the inside – so we count angles greater than $180^\circ$ as such.

A fan triangulation from a single vertex may fail – some diagonals can pass outside the polygon. It is still true, however, that every simple $n$-gon can be split into $n-2$ triangles (not necessarily as a fan), which rescues the proof. The reason is induction via a triangle formed by some vertex $V$ and its neighbors $U$, $W$ that lies entirely inside our $n$-gon. Cutting $V$ off along the diagonal $UW$ gives an $(n-1)$-gon; iterating yields $n-2$ triangles. One can prove that such a triangle always exists (in fact, two of them); see the two ears theorem.

As for Proof 3: it works only if there is an interior point $O$ visible from every vertex (i.e., the segments $OA_i$ lie entirely inside). For some non-convex polygons no such $O$ exists (try drawing one) – in that case Proof 3 cannot be salvaged directly and one must fall back on the triangulation described in the previous paragraph.

Forward implication. Assume $AB = AC$. Let $M$ be the midpoint of $BC$. The triangles $ABM$ and $ACM$ are congruent by $SSS$ ($AB=AC$ by assumption, $BM=CM$ by definition of the midpoint, common side $AM$), from which $\angle ABC = \angle ACB$. This proof also works with other choices of $M$. If we take $M$ to be the foot of the perpendicular from $A$ to $BC$, we obtain the common side $AM$, $AB=AC$ and a right angle at $M$, which lies opposite the longest side (the hypotenuse) $AB$, resp. $AC$ – we apply $SsA$. If we take $M$ to be the foot of the bisector of the interior angle at $A$, we have $AB=AC$, the equal angle $\angle BAM=\angle CAM$, and the common side $AM$ between them – we apply $SAS$.

Converse implication. Assume $\angle ABC = \angle ACB = \beta$. Let $M$ be the foot of the perpendicular from $A$ to $BC$. By $ASA$, $\triangle ABM \cong \triangle ACM$, since we have two equal angles $\angle ABM = \angle ACM$ and $\angle AMB = \angle AMC = 90^\circ$ and the common side $AM$, so $AB = AC$. Similarly, we could take $M$ to be the foot of the bisector of the interior angle at $A$ – from the equal angles $\angle BAM=\angle CAM$ and $\angle ABM=\angle ACM$ and the common side $AM$ we again apply $ASA$. Taking $M$ to be the midpoint of $BC$, however, doesn't help here: we get $BM=CM$, $AM$ common, and $\angle ABM=\angle ACM=\beta$, but the angle $\beta$ lies opposite $AM$, which may be shorter than $BM$ (when the angle at $A$ is obtuse), so even $SsA$ cannot be applied in general – the lesson is that even with the right point, its precise definition matters (we will see this many more times).

Poznámka. Another cute proof is based on showing that $\triangle ABC \cong \triangle ACB$ (note the different order of vertices). In the case where we know the equal sides, we use $SAS$ or even $SSS$ – which then gives equal angles. In the case where we know the equal angles, it will instead be $ASA$, and the congruence gives equal sides.

($\Rightarrow$) Let point $X$ lie on the perpendicular to $AB$ through the midpoint $M$. The triangles $AMX$ and $BMX$ are congruent by $SAS$: $MA=MB$ (midpoint), common side $MX$, and equal right angles at $M$. Hence $XA=XB$.

($\Leftarrow$) Suppose point $X$ satisfies $XA=XB$ but does not lie on the perpendicular to $AB$ through $M$. If $X$ lies on the line $AB$, it is clear that the only point of this line equidistant from $A$ and $B$ is precisely the midpoint $M$.

Suppose then that $X$ does not lie on the line $AB$. Without loss of generality, let our perpendicular be vertical and let $X$ lie to its left. Denote by $Y$ the intersection of segment $XB$ with this perpendicular. By the already proved implication, $YA=YB$.

In the isosceles $\triangle XAB$ ($XA=XB$) we have $\angle XAB=\angle XBA$. Since $Y$ lies on segment $XB$, the rays $BX$ and $BY$ coincide, so $\angle XBA=\angle YBA$. But in the isosceles $\triangle YAB$ ($YA=YB$) we have $\angle YBA=\angle YAB$. Altogether,

This, however, is nonsense: point $Y$ lies inside segment $XB$, so the ray $AY$ points into the interior of triangle $XAB$ and lies strictly inside the angle $\angle XAB$, i.e. $\angle YAB<\angle XAB$.

Let $ABC$ be a triangle. The perpendicular bisector of segment $AB$ is perpendicular to $AB$, and the perpendicular bisector of segment $AC$ is perpendicular to $AC$; these two perpendiculars are parallel only when $AB \parallel AC$, which clearly does not hold. Hence the perpendicular bisectors of segments $AB$ and $AC$ meet at a single point – call it $O$.

By the previous theorem, $OA=OB$ (since $O$ lies on the perpendicular bisector of $AB$) and $OA=OC$ (since $O$ lies on the perpendicular bisector of $AC$). Together this gives $OB=OC$, which in turn tells us that $O$ also lies on the perpendicular bisector of segment $BC$. All three perpendicular bisectors therefore pass through $O$.

Poznámka (1). The point $O$ satisfies $OA=OB=OC$, so it is the center of a circle passing through all three vertices – that is, the circumscribed circle (circumcircle) of triangle $ABC$.

Poznámka (2). The notation $O$ for the circumcenter is the international standard. In Czechia and Slovakia, the notation $S$ is also used (from the word střed, meaning center).

($\Rightarrow$) Let the ray $AZ$ split the angle $XAY$ into two equal angles. Let $X'$, $Y'$ be the feet of the perpendiculars dropped from $Z$ to the lines $AX$, $AY$. The right triangles $AX'Z$ and $AY'Z$ are congruent by $ASA$: common hypotenuse $AZ$, equal angles at $A$, and equal right angles at $X'$, $Y'$. Hence $ZX'=ZY'$, which are precisely the distances from $Z$ to the lines $AX$ and $AY$.

($\Leftarrow$) Conversely, suppose $ZX'=ZY'$, where $X'$, $Y'$ are again the feet of the perpendiculars from $Z$ to the lines $AX$, $AY$. The right triangles $AX'Z$ and $AY'Z$ are congruent by $SsA$: common hypotenuse $AZ$, equal legs $ZX'=ZY'$, and equal right angles at $X'$, $Y'$ opposite this hypotenuse (the longest side). Hence $\angle X'AZ=\angle Y'AZ$, which is precisely $\angle XAZ=\angle ZAY$.

Let $ABC$ be a triangle. Let $I$ be the intersection of the internal angle bisectors at vertices $B$ and $C$, which clearly exists and clearly lies inside $ABC$.

By the previous theorem, $d(I, AB) = d(I, BC)$ (since $I$ lies on the bisector of the angle at $B$) and $d(I, BC) = d(I, AC)$ (since $I$ lies on the bisector of the angle at $C$). Hence $d(I, AB) = d(I, AC)$. Since $I$ lies inside the angle $BAC$, this means that $I$ lies on its bisector. All three internal bisectors therefore pass through $I$.

Poznámka (1). The point $I$ is equidistant from all three sides, so it is the center of a circle tangent to all three sides from the inside – the incircle of triangle $ABC$.

Poznámka (2). The notation $I$ comes from the word incenter. In the Czech-Slovak olympiad the letter $S$ is often used instead (for střed, meaning center).

For clarity of notation, denote by $X$, $Y$ points on the rays opposite to $CA$, $BA$. The external bisector of the angle at $B$ is exactly the internal bisector of the angle $YBC$, and the external bisector at $C$ is the internal bisector of the angle $XCB$. Let $I_a$ be the intersection of these bisectors.

From the angle bisector theorem applied to the angles $YBC$ and $XCB$ we obtain the equalities

The point $I_a$ is therefore equidistant from the lines $AB$ and $AC$, so it lies on either the internal or the external bisector of the angle $BAC$. But it clearly lies inside the angle $BAC$, so it is on the internal bisector – which therefore passes through $I_a$.

Poznámka (1). The point $I_a$ is equidistant from the line $BC$ and from the lines containing the sides $AB$, $AC$, so it is the center of a circle tangent to the side $BC$ externally and to the other two sides (more precisely their extensions) internally – the excircle of the triangle opposite vertex $A$.

Poznámka (2). The notation $I_a$ is the common international notation. The escribed circle is called the excircle, so perhaps $E_a$ would have been more logical, but it is rarely used. Then again, the inscribed circle is associated with the incenter, so perhaps that's why 🙃.