S is any point in the interior of ∆PQR. Show that ABCD AB + BC + CD + DA > AC + BD.
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Solution

Produce QS so that it meets PR at T.

$In △ P Q T, we have : P Q + P T > Q T \Rightarrow P Q + P T > S Q + S T . . . (i) In △ S T R, we have : S T + T R > S R . . . . (i i) Adding (i) and (ii), we get : P Q + P T + S T + T R > S Q + S T + S R \Rightarrow P Q + P T + T R > S Q + S R \Rightarrow P Q + P R > S Q + S R \Rightarrow S Q + S R < P Q + P R$

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In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that

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