Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
In Constructing Perpendicular and Parallel Lines you constructed perpendicular bisectors and said they hold the points that are equally far from both endpoints. Now we prove it with Proofs with Congruent Triangles, and use it to find the one point that is equally far from all three vertices of a triangle.
💡 Por qué funciona
Perpendicular Bisector Theorem: if $P$ is on the perpendicular bisector of $\overline{AB}$, then $PA = PB$. If $M$ is the midpoint, the triangles $PMA$ and $PMB$ share $\overline{PM}$, have right angles at $M$ and $MA = MB$, so they are congruent by SAS and $PA = PB$. Converse: if $PA = PB$, then $P$ is on the perpendicular bisector (the two right triangles are congruent by SSS, so the angles at $M$ are equal and therefore right).
The circumcenter. Let $O$ be where the perpendicular bisectors of $\overline{AB}$ and $\overline{BC}$ cross. Then $OA = OB$ and $OB = OC$, so $OA = OC$, which puts $O$ on the third bisector too. All three bisectors meet at $O$, and the circle centered at $O$ through $A$ passes through all three vertices: the circumscribed circle.
The three perpendicular bisectors meet at the circumcenter $O$. The red radii $OA$, $OB$ and $OC$ are equal.
🔑 Idea clave
A point is on the perpendicular bisector of $\overline{AB}$ exactly when it is equally far from $A$ and $B$. The three perpendicular bisectors of a triangle meet at the circumcenter, the center of the circle through all three vertices.
✏️ Ejemplo resuelto: A point on a bisector
$P$ is on the perpendicular bisector of $\overline{AB}$, $PA = 4x - 3$ and $PB = 2x + 9$. Find $PA$.
$PA = PB$: $4x - 3 = 2x + 9$, so $2x = 12$ and $x = 6$.
$PA = 4(6) - 3 = 21$.
Respuesta: $PA = 21$
✏️ Ejemplo resuelto: The circumcenter on a grid
Find the circumcenter of $A(0, 0)$, $B(6, 0)$ and $C(2, 4)$.
$\overline{AB}$ is horizontal with midpoint $(3, 0)$, so its perpendicular bisector is $x = 3$.
The circumcenter is $(3, y)$ with $OA = OC$: $3^2 + y^2 = (3 - 2)^2 + (y - 4)^2$.
$9 + y^2 = 1 + y^2 - 8y + 16$, so $8y = 8$ and $y = 1$.
Where the circumcenter lands depends on the largest angle: inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, and outside an obtuse triangle.
⚠️ Error común
Thinking the circumcenter is equally far from the sides. It is equally far from the three vertices. (The point equally far from the sides is the incenter, in Incenter, Centroid and Orthocenter.)
Práctica
Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 2 XP.