In an equilateral triangle A B C, D is a point on the side BC such that BD =1/3BC. Prove that 9 AD²=7 AB².

OR

Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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