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प्रश्न
ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2= AC2 + BD2
उत्तर १
Let the diagonals AC and BD of rhombus ABCD intersect at O.
Since the diagonals of a rhombus bisect each other at right angles.
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90º and AO = CO, BO = OD.
Since ∆AOB is a right triangle right-angle at O.
∴ AB2 = OA2 + OB2
`AB^2=(1/2AC)^2 +(1/2BD)^2 `
⇒ 4AB2 = AC2 + BD2 ….(i)
Similarly, we have
`4BC^2 = AC^2 + BD^2 ….(ii)`
`4CD^2 = AC^2 + BD^2 ….(iii)`
and,
`4AD^2 = AC^2 + BD^2 ….(iv)`
Adding all these results, we get
`4(AB^2 + BC^2 + AD^2 ) = 4(AC^2 + BD^2 )`
`⇒ AB^2 + BC^2 + AD^2 + DA^2 = AC^2 + BD^2`
उत्तर २
In ΔAOB, ΔBOC, ΔCOD, ΔAOD
Applying Pythagoras theorem
AB2 = AO2 + OB2
BC2 = BO2 + OC2
CD2 = CO2 + OD2
AD2 = AO2 + OD2
Adding all these equations,
AB2 + BC2 + CD2 + AD2
= 2(AO2 + OB2 + OC2 + OD2)
= `2{("AC"/2)^2 + ("BD"/2)^2 + ("AC"/2)^2 + ("BD"/2)^2}` ...(diagonals bisect each other)
= `2(("AC")^2/2 + ("BD")^2/2)`
= (AC)2 + (BD)2
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