Answer :

Given: AB = BC = CA = x (say)

BD = BC


To prove: 9 AD2 = 7 AB2


Proof: Construct AE perpendicular to BC.



As BC = x,


BD =


In ∆ABE and ∆ACE,


AB = AC [ AB = AC = x]


AE=AE [common sides in both triangles]


AEB = AEC [∵∠AEB = AEC = 90°]


Thus, ∆ABE ∆ACE by RHS congruency, i.e., Right angle-Hypotenuse-Side congruency.


If ∆ABE ∆ACE,


BE = CE


[ corresponding parts of congruent triangles are congruent]


So, BE = CE = BC


BE = CE =


We have BE = x/2, BD = x/3 and clearly


BD + DE = BE




In ∆ABE using Pythagoras theorem,


(hypotenuse)2 = (perpendicular)2 + (base)2


x2 = AE2 +


AE2 = x2


AE2 = …(i)


Similarly, using pythagoras theorem in right ∆AED,


AD2 = AE2 + ED2


AD2 = +


AD2 =


AD2 =


AD2 =


9 AD2=7 x2


9AD2 =7 AB2 [ AB = x AB2 = x2]


OR


Need to prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals



ABCD is a rhombus in which diagonals AC and BD intersect at point O.


We need to prove AB2 + BC2 + CD2 + DA2 = AC2 + DB2


In Δ AOB; AB2 = AO2 + BO2


In Δ BOC; BC2 = CO2 + BO2


In Δ COD; CD2 = DO2 + CO2


In Δ AOD; AD2 = DO2 + AO2


Adding the above 4 equations we get


AB2 + BC2 + CD2 + DA2


= AO2 + BO2 + CO2 + BO2 + DO2 + CO2 + DO2 + AO2


= 2(AO2 + BO2 + CO2 + DO2)


Since, AO2 = CO2 and BO2 = DO2 = 2(2 AO2 + 2 BO2)


= 4(AO2 + BO2) ……eq(1)


Now, let us take the sum of squares of diagonals


AC2 + DB2 = (AO + CO)2 + (DO+ BO)2 = (2AO)2 + (2DO)2


= 4 AO2 + 4 BO2 ……eq(2)


From eq(1) and eq(2) we get


AB2 + BC2 + CD2 + DA2 = AC2 + DB2


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