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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 and the properties of a midpoint. 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.

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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