If o is any point inside the quadrilateral abcd then prove that oa ob oc odac bd?

Answer 1

Well, darling, let me break it down for you. If point O is inside quadrilateral ABCD, then the sum of the distances from point O to each of the four sides of the quadrilateral is equal to the sum of the diagonals AC and BD. It's a little math magic that shows O is cozy in the middle of the quadrilateral, no matter which way you slice it.

Answer 2

To prove: OA + OB + OC+OD > AC + BD

Construction:Join OA, OB, OC and OD. Also, join AC and BD

Proof : Consider ΔBOD, by triangle in equality (sum of any two sides of a triangle is greater than the third side),

We have OB + OD >BD .....(1)

Similarly, in ΔAOC, by triangle inequality, OA + OC > AC .... (2)

Adding (1) and (2)

(OB + OD) + OA + OC ) > BD + AC

⇒ OA + OB + OC + OD > AC + BD

Hence, the result is proved.

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OR

Here is the detailed explanation.

Let O be any point within the quadrilateral ABCD.

To prove: OA + OB + OC > AC + BD

Construction:Join OA, OB, OC and OD. Also, join AC and BD

Proof : Consider ΔBOD, by triangle in equality (sum of any two sides of a triangle is greater than the third side),

We have OB + OD >BD .....(1)

Similarly, in ΔAOC, by triangle inequality, OA + OC > AC .... (2)

Adding (1) and (2)

(OB + OD) + OA + OC ) > BD + AC

⇒ OA + OB + OC + OD > AC + BD

Hence, the result is proved.