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Figures on the Same Base and Between the Same Parallels

𝐎 is a point...

Question

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

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Solution

please re-check the question.
Question will be
OA+OB+OC+OD>AC+BD.

We cant connect BC in the question.

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Similar questions

O A + O B + O C + O D > A C + B D

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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