Q. 35

# In an equilateral

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,

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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