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\(ABCD\) is quadrilateral. Is \(AB + BC + CD + DA < 2 (AC + BD)\)
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Question

\(ABCD\) is quadrilateral. Is \(AB + BC + CD + DA < 2 (AC + BD)\)

Answer

Yes
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Solution

Let us consider \(ABCD\) is quadrilateral and \(P\) is the point where the diagonals are intersect. As shown in the figure below.
Solution for ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)
We know that,
The sum of the length of any two sides is always greater than the third side.
Now consider the \(ΔPAB\),
Here, \(PA+PB>AB\) … [equation i]
Then, consider the \(ΔPBC\)
Here, \(PB+PC>BC\) … [equation ii]
Consider the \(ΔPCD\)
Here, \(PC+PD>CD\) … [equation iii]
Consider the \(ΔPDA\)
Here, \(PD+PA>DA\) … [equation iv]
By adding equation [i], [ii], [iii] and [iv], we get,\(PA+PB+PB+PC+PC+PD+PD+PA>AB+BC+CD+DA\)
\(\implies 2PA+2PB+2PC+2PD>AB+BC+CD+DA\)
\(\implies 2PA+2PC+2PB+2PD>AB+BC+CD+DA\)
\(\implies 2(PA+PC)+2(PB+PD)>AB+BC+CD+DA\)
From the figure we have, \(AC = PA + PC\) and \(BD = PB + PD\)
Then,
\(2AC+2BD>AB+BC+CD+DA\)
\(\implies 2(AC+BD)>AB+BC+CD+DA\)
Hence, the given expression is true.
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