ABCD is a rhombus. Prove that AB square+Bc square+CD square+Ad square = AC square+ BD square
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ABCD is a rhombus. Prove that AB2+BC2+CD2+DA2=AC2+BD2 (u may copy this question)
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Given,
ABCD is a rhombus.
To find,
Prove AB² + BC² + CD² + AD² = AC² + BD²
Solution,
We know that ABCD is a rhombus.
Therefore, AB = BC= CD = DA and ∠A = ∠B = ∠C = ∠D = 90°
In Δ AOD,
It is a right angle triangle. By Pythagoras theorem,
AD² = A0² + OD² (i)
Similarly, in Δ COD
CD² = CO² + OD² (ii)
Similarly, in Δ COB
BC² = CO² + OB² (iii)
Similarly, in Δ AOB
AB² = AO² + OB² (iv)
Adding equation (i), (ii), (iii) and (iv)
AD² + CD² + BC² + AB² = 2 AO² + 2 CO² + 2 OD² + 2 OB²
AD² + CD² + BC² + AB² = 2 (AO² + CO²) + 2 (OB² + OD²)
But since AO = CO = AC/2, and OB = OD = BD/2
AD² + CD² + BC² + AB² = 2 [(AC/2)² + (AC/2)²] + 2 [(BD/2)² + (BD/2)²]
AD² + CD² + BC² + AB² = 2 [2AC²/4] + 2[2BD²/4]
AD² + CD² + BC² + AB² = AC² + BD²
AD² + CD² + BC² + AB² = AC² + BD²
Hence proved that AB² + BC² + CD² + AD² = AC² + BD²
