Ibrahim Rosales

2022-07-20

Point $A$ has coordinates $\left({x}_{1},{y}_{1}\right)$, point $B$ has coordinates $\left({x}_{2},{y}_{2}\right)$ andpoint $M$ has coordinates $\left(\frac{{x}_{1}+{x}_{1}}{2},\frac{{y}_{1}+{y}_{1}}{2}\right)$
Prove that $AM=\frac{1}{2}AB$

decoratesuw

Expert

The points $A\left({x}_{1},{y}_{1}\right),B\left({x}_{2},{y}_{2}\right),M\left({x}_{m}=\frac{{x}_{1}+{x}_{2}}{2},{y}_{m}=\frac{{y}_{1}+{y}_{2}}{2}\right)$ are given.
First note that the points $A$, $M$ and $B$ are colinear (i.e. lie on the same line):
$\text{slope of AM}=\frac{{y}_{m}-{y}_{1}}{{x}_{m}-{x}_{1}}=\frac{\frac{{y}_{1}+{y}_{2}}{2}-{y}_{1}}{\frac{{x}_{1}+{x}_{2}}{2}-{x}_{1}}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\text{slope of AB}.$
The triangles $\mathrm{\Delta }AMD$ and $\mathrm{\Delta }ABC$ are similar, because their corresponding angles are equal. Hence:
$\frac{AM}{AB}=\frac{AD}{AC}=\frac{AD}{2AD}=\frac{1}{2}⇒AM=\frac{1}{2}AB.$

Taniya Burns

Expert

$\begin{array}{rl}A{M}^{2}& ={\left(\frac{{x}_{1}+{x}_{2}}{2}-{x}_{1}\right)}^{2}+{\left(\frac{{y}_{1}+{y}_{2}}{2}-{y}_{1}\right)}^{2}\\ & ={\left(\frac{-{x}_{1}+{x}_{2}}{2}\right)}^{2}+{\left(\frac{-{y}_{1}+{y}_{2}}{2}\right)}^{2}\\ & =\frac{1}{4}\left(\left(-{x}_{1}+{x}_{2}{\right)}^{2}+\left(-{y}_{1}+{y}_{2}{\right)}^{2}\right)=\frac{1}{4}A{B}^{2},\end{array}$
from which it follows that $AM=\frac{1}{2}AB$.

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