Difference between revisions of "Midpoint"

Revision as of 22:00, 24 February 2021

Definition

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

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);
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Midsegments

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 $\begin{align*} [ABC] &= 4[EFD] \end{align*}$

Medians

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: $\begin{align*} M(\frac{x_A+x_B}{2}&,\frac{y_A+y_B}{2}) \end{align*}$

@@ Line 1: / Line 1: @@
 == Definition ==
-The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.
+In '''Euclidean geometry''', the '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.
 A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment <math>\overline{AB}</math> with midpoint <math>M</math>, <math>AM=BM=\frac{1}{2}AB</math>. Alternatively, any point <math>M</math> on <math>\overline{AB}</math> such that <math>AM=BM</math> is the midpoint of the segment.
@@ Line 17: / Line 17: @@
 pair A,B,C,D,E,F,G;
 A=(0,0);
-B=(4,0);
+B=(4,0;
-C=(1,3);
+C=(1,3)
 D=(2,0);
 E=(2.5,1.5);

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