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Difference between revisions of "Midpoint"
Line 12: Line 12:
 
dot((2,0));
 
dot((2,0));
 
label("M",(2,0),N);
 
label("M",(2,0),N);
 +
label("Figure 1",(2,0),4S);
 
</asy>
 
</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>
 +
=== Midsegments ===
 +
As shown in Figure 2, <math>\Delta ABC</math> is a triangle with <math>D</math>, <math>E</math>, <math>F</math> midpoints on <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CA}</math> respectively. Connect <math>\overline{EF}</math>, <math>\overline{FD}</math>, <math>\overline{DE}</math> (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that <math>\Delta CFE \sim \Delta CAB</math> and likewise for <math>\Delta ADF</math> and <math>\Delta BED</math>. Because of this, we know that
 +
<cmath>\begin{align*}
 +
AB &= 2FE \\
 +
BC &= 2DE \\
 +
CA &= 2ED \\
 +
\end{align*}</cmath>
 +
Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, <math>\Delta ABC \sim \Delta EFD (SSS)</math> with similar ratio 2:1. The area ratio is then 4:1; this tells us
 +
<cmath>\begin{align*}
 +
[ABC] &= 4[EFD]
 +
\end{align*}</cmath>
 
== In Cartesian Plane ==
 
== In Cartesian Plane ==
 
In the Cartesian Plane, the coordinates of the midpoint <math>M</math> can be obtained when the two endpoints <math>A</math>, <math>B</math> of the line segment <math>\overline{AB}</math> is known. Say that <math>A: A(x_A,y_A)</math> and <math>B: B(x_B,y_B)</math>. The Midpoint Formula states that the coordinates of <math>M</math> can be calculated as:
 
In the Cartesian Plane, the coordinates of the midpoint <math>M</math> can be obtained when the two endpoints <math>A</math>, <math>B</math> of the line segment <math>\overline{AB}</math> is known. Say that <math>A: A(x_A,y_A)</math> and <math>B: B(x_B,y_B)</math>. The Midpoint Formula states that the coordinates of <math>M</math> can be calculated as:

Revision as of 02:31, 12 February 2021

This article is a stub. Help us out by expanding it.

Definition

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]

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*}

In 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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