Question

ABCD is a rhombus. Prove that AB square+Bc square+CD square+Ad square = AC square+ BD square​

Answer 1

ABCD  is a rhombus. Prove that AB2+BC2+CD2+DA2=AC2+BD2 (u may copy this question)  

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Answer 2

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