Show that AB + BC + CD + DA > AC + BD
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Step-by-step explanation:
Answer. ABCD is a quad. Its diagonals are AC and BD. AB + BC + CD + AD > AC + BD.
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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)
HENCE, PROVED
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