Answer :

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

BD = BC

To prove: 9 AD^{2} = 7 AB^{2}

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}

⇒ x^{2} = AE^{2} +

⇒ AE^{2} = x^{2} –

⇒ AE^{2} = …(i)

Similarly, using pythagoras theorem in right ∆AED,

AD^{2} = AE^{2} + ED^{2}

⇒ AD^{2} = +

⇒ AD^{2} =

⇒ AD^{2} =

⇒ AD^{2} =

⇒ 9 AD^{2}=7 x^{2}

⇒ 9AD^{2} =7 AB^{2} [∵ AB = x ⇒ AB^{2} = x^{2}]

**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 AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + DB^{2}

⇒ In Δ AOB; AB^{2} = AO^{2} + BO^{2}

⇒ In Δ BOC; BC^{2} = CO^{2} + BO^{2}

⇒ In Δ COD; CD^{2} = DO^{2} + CO^{2}

⇒ In Δ AOD; AD^{2} = DO^{2} + AO^{2}

⇒ Adding the above 4 equations we get

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2}

= AO^{2} + BO^{2} + CO^{2} + BO^{2} + DO^{2} + CO^{2} + DO^{2} + AO^{2}

⇒ = 2(AO^{2} + BO^{2} + CO^{2} + DO^{2})

Since, AO^{2} = CO^{2} and BO^{2} = DO^{2} = 2(2 AO^{2} + 2 BO^{2})

= 4(AO^{2} + BO^{2}) ……eq(1)

Now, let us take the sum of squares of diagonals

⇒ AC^{2} + DB^{2} = (AO + CO)^{2} + (DO+ BO)^{2} = (2AO)^{2} + (2DO)^{2}

= 4 AO^{2} + 4 BO^{2} ……eq(2)

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

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + DB^{2}

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