Triangle Inequality

This week, we learn about the Triangle Inequality, one of the simplest geometric inequalities with many consequences.

How would you use Triangle Inequality to solve the following?

Given quadrilateral $ABCD$ , let $E$ and $F$ be the midpoints of $AD$ and $BC$ respectively. Show that $AB+DC \geq 2 EF.$ Can equality hold? If yes, when?

#Geometry #KeyTechniques #Inequality

Easy Math Editor

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Comments

Let A,B,C,D,E,F have position vectors $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d},\mathbf{e},\mathbf{f}$ . Then $\begin{array}{rcl} \overrightarrow{AB} & = & \mathbf{b} - \mathbf{a} \\ \overrightarrow{DC} & = & \mathbf{c} - \mathbf{d} \\ \overrightarrow{EF} & = & \mathbf{f} - \mathbf{e} \; = \; \tfrac12(\mathbf{b}+\mathbf{c}) - \tfrac12(\mathbf{a}+\mathbf{d}) \\ & = & \tfrac12(\mathbf{b} - \mathbf{a}) + \tfrac12(\mathbf{c} - \mathbf{d}) \\ 2\overrightarrow{EF} & = & \overrightarrow{AB} + \overrightarrow{DC} \end{array}$ By the triangle inequality, then $\begin{array}{rcl} 2 EF & = & 2\big|\overrightarrow{EF}\big| \; = \; \big|\overrightarrow{AB} + \overrightarrow{DC}\big| \; \le \; \big|\overrightarrow{AB}\big| + \big|\overrightarrow{DC}\big| \; = \; AB + DC \end{array}$ with equality when $AD$ and $BC$ are parallel (they can't be antiparallel), so $ABCD$ is a parallelogram or a trapezium.

Mark Hennings - 7 years, 8 months ago

Let P be th midpoint of DB.

$FP=AB/2$ and $EP=DC/2$

From Triangle Inequality on triangle $FPE$ ,

we get $FP +EP =AB/2 + DC/2$ > $FE$

OR $AB +DC >2 \times FE$

Equatlity holds if and only if $P$ falls on $FE$

Or in other words $FE$ , $FP$ and $EP$ coincide.

Ranjana Kasangeri - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$