Question

Show that in a quadrilateral abcd ab+bc+cd+da 2(bd+ac)

Answer 1

Answer:

ABCD is a quadrilateral and 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

Step-by-step explanation:

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

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

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > AC

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > AC

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BD

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BDBC + CD > BD

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BDBC + CD > BDAdding all the above equations,

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BDBC + CD > BDAdding all the above equations,2(AB + BC + CA + AD) > 2(AC + BD)

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BDBC + CD > BDAdding all the above equations,2(AB + BC + CA + AD) > 2(AC + BD)⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BDBC + CD > BDAdding 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)

Sum of the two sides of a triangle is greater than the third side.So, considering the triangle ABC, BCD, CAD and BAD, we getAB + BC > ACCD + AD > ACAB + AD > BDBC + CD > BDAdding 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