Question
In a quadrilateral $ABCD$, show that $(AB + BC + CD + DA) > (AC + BD)$.
✓
Answer
Given: Quadrilateral $ABCD$
To prove: $(AB + BC + CD + DA) > (AC + BD)$
Proof:
In $\triangle\text{ABC},$ $\text{AB + BC > AC }...(\text{i})$ In $\triangle\text{CAD},$
$\text{CD + AD > AC }...(\text{ii})$ In $\triangle\text{BAD,}$
$\text{AB + AD + BD } ...(\text{iii})$ In $\triangle\text{BCD},$
$\text{BC + CD + BD }...(\text{iv})$ Adding $(i), (ii), (iii)$ and $(iv)$,
we get $2(\text{AB + BC + CD + DA})< 2 \text{(AC + BD)}$
Hence, $\text{(AB + BC + CD + DA)}<(\text{AC + BD}).$
To prove: $(AB + BC + CD + DA) > (AC + BD)$
Proof:
In $\triangle\text{ABC},$ $\text{AB + BC > AC }...(\text{i})$ In $\triangle\text{CAD},$
$\text{CD + AD > AC }...(\text{ii})$ In $\triangle\text{BAD,}$
$\text{AB + AD + BD } ...(\text{iii})$ In $\triangle\text{BCD},$
$\text{BC + CD + BD }...(\text{iv})$ Adding $(i), (ii), (iii)$ and $(iv)$,
we get $2(\text{AB + BC + CD + DA})< 2 \text{(AC + BD)}$
Hence, $\text{(AB + BC + CD + DA)}<(\text{AC + BD}).$
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