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?

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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. – Mark Hennings · 3 years, 10 months ago

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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. – Ranjana Kasangeri · 3 years, 10 months ago

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