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In Euclidean geometry, the midpoint of a line segment is the point on the segment equidistant from both endpoints.

A midpoint bisects the line segment that the midpoint lies on. Because of this property, we say that for any line segment $\overline{AB}$ with midpoint $M$, $AM=BM=\frac{1}{2}AB$. Alternatively, any point $M$ on $\overline{AB}$ such that $AM=BM$ is the midpoint of the segment. [asy] draw((0,0)--(4,0)); dot((0,0)); label("A",(0,0),N); dot((4,0)); label("B",(4,0),N); dot((2,0)); label("M",(2,0),N); label("Figure 1",(2,0),4S); [/asy]

Midpoints and Triangles

[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(4,0); C=(1,3); D=(2,0); E=(2.5,1.5); F=(0.5,1.5); G=(5/3,1); draw(A--B--C--cycle); draw(D--E--F--cycle,green); dot(A--B--C--D--E--F--G); draw(A--E,red); draw(B--F,red); draw(C--D,red); label("A",A,S); label("B",B,S); label("C",C,N); label("D",D,S); label("E",E,E); label("F",F,W); label("G",G,NE); label("Figure 2",D,4S); [/asy]


As shown in Figure 2, $\Delta ABC$ is a triangle with $D$, $E$, $F$ midpoints on $\overline{AB}$, $\overline{BC}$, $\overline{CA}$ respectively. Connect $\overline{EF}$, $\overline{FD}$, $\overline{DE}$ (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that $\Delta CFE \sim \Delta CAB$ and likewise for $\Delta ADF$ and $\Delta BED$. Because of this, we know that \begin{align*} AB &= 2FE \\ BC &= 2DE \\ CA &= 2ED \\ \end{align*} Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, $\Delta ABC \sim \Delta EFD  (SSS)$ with similar ratio 2:1. The area ratio is then 4:1; this tells us \begin{align*} [ABC] &= 4[EFD] \end{align*}


The median of a triangle is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are $\overline{AE}$, $\overline{BF}$, and $\overline{CD}$, segments highlighted in red.

These three line segments are concurrent at point $G$, which is otherwise known as the centroid. This concurrence can be proven through many ways, one of which involves the most simple usage of Ceva's Theorem. A median is always within its triangle.

The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.

For right triangles, the median to the hypotenuse always equals to half the length of the hypotenuse.

For equilateral triangles, its median to one side is the same as the angle bisector and altitude. It can be calculated as $\frac{\sqrt3}{2}s$, where $s$ denotes its side length.

Cartesian Plane

In the Cartesian Plane, the coordinates of the midpoint $M$ can be obtained when the two endpoints $A$, $B$ of the line segment $\overline{AB}$ is known. Say that $A: A(x_A,y_A)$ and $B: B(x_B,y_B)$. The Midpoint Formula states that the coordinates of $M$ can be calculated as: \begin{align*} M(\frac{x_A+x_B}{2}&,\frac{y_A+y_B}{2}) \end{align*}

See Also

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