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Theorem 2: Triangles

If O is a poi...

Question

If O is a point within a quadrilateral ABCD, prove that ; $O A + O B + O C + O D > A C + B D$ .

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Solution

In $Δ O A C$
$O A + O C > A C$ _ (1)
(sum of two sides of a $Δ$ is greater than third side)
similarity in $Δ O B D$
$O B + O D > B D$ ___ (2)
(1)+(2)
$O A + O B + O C + O D > A C + B D$

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Similar questions

O is a point inside a quadrilateral ABCD which isn't at the point of intersection of diagonals. Prove that

OA+OB+OC+OD>AC+BC

A D = B C

A C

B D

O

\frac{O C}{O A} = \frac{O D}{O B} = \frac{1}{3}

A B C D

A B C D

A C

B D

O

\frac{O C}{O A} = \frac{O D}{O B} = \frac{1}{3}

△ O A B \sim △ O C D

A O = 2 C O

B O = 2 D O

O A \times O D = O B \times O C

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