In a Quadrilateral Abcd, ∠B = 90° And ∠D = 90°. Prove That: 2ac2 - Ab2 = Bc2 + Cd2 + Da2 - Mathematics

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Sum

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC2 - AB2 = BC2 + CD2 + DA2

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Solution


In quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
So, ΔABC and ΔADC are right-angled triangles.

In ΔABC, using Pythagoras theorem,
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2                  ....(i)

In ΔADC, using Pythagoras theorem,
AC2 = AD2 + DC2                   ....(ii)

LHS = 2AC2 - AB2
= 2AC2 - ( AC2 - BC2 )           .....[ From(i) ]
= 2AC2 - AC2 + BC2
= AC2 + BC2
= AD2 + DC2 + BC           ....[ From(ii) ]
= RHS

  Is there an error in this question or solution?
Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [Page 164]

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Selina Concise Mathematics Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 7 | Page 164

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