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?

No vote yet

17 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet 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.

Log in to reply

Let

Pbe th midpoint ofDB.\(FP=AB/2\) 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.

Log in to reply