Question

ABCD is a quadrilateral prove that (AB+BC+CD+DC)> (AC+BD)
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Answer 1

Answer:

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Step-by-step explanation:

ABCD is a quadrilateral. Joining A to C and B to D, we get triangles, ABC, BDC, ACD and ABD. AC and BD are the diagonals.

Sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

⇒ (AB + BC + CA + AD) > (AC + BD)

⇒ (AC + BD) < (AB + BC + CA + AD)

Answer 2

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here is the answer.....

ABCD is a quadrilateral. Joining A to C and B to D, we get triangles, ABC, BDC, ACD and ABD. AC and BD are the diagonals.

Sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

⇒ (AB + BC + CA + AD) > (AC + BD)

⇒ (AC + BD) < (AB + BC + CA + AD)

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