In an equilateral ∆ABC, D is a point on side BC such that . Prove that 9(AD)² = 7(AB)²

OR

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Given:

• Equilateral Triangle ABC with side say “x”units

•

Now, lets construct AE ⊥ BC

In ∆ AEC and ∆ AEB

• AC = AB = x

• ∠ AEC = ∠ AEB = 90°

• AE = AE (common side)

So, by RHS rule, ∆ AEC ≅ ∆ AEB

∴ CE = BE

(∵ Corresponding sides of congruent triangles are equal)

As given in the question

So,

Using Pythagoras Theorem,

AE^{2 =} AC^{2 –} EC²

Similarly, AD^{2 =} AE² + ED²

And, AB² = x²

∴

Hence, Proved

OR

Given:

• Square ABCD with side (a + b) units

• Points E, F, G, H on the sides such that they divide the segment in the same proportion.

Now,

In ∆ EAF and ∆ HBE,

∠ A = ∠ B

EA = HB = b

AF = BE = a

So, by SAS rule, ∆ EAF ≅ ∆ HBE

Similarly, ∆ HBE ≅ ∆ GCH and ∆ GCH ≅ ∆ FDG

∴ ∆ EAF ≅ ∆ HBE ≅ ∆ GCH ≅ ∆ FDG

EF = FG = GH = HE = c

(∵ corresponding sides of congruent triangles are equal)

Also, ∠ AEF = ∠ BHE, which implies,
∠ AEF + ∠ BEH + ∠ FEH = 180°

(Straight line measures 180°)
⇒ (∠ BHE + ∠ BEH) + ∠ FEH = 180°

(∵ ∠ BHE = ∠ AEF)

⇒ ∠ FEH = 180° – (∠ BHE + ∠ BEH)

= 180° – 90° = 90°

(∵ ∆ BEH is a Right Triangle)

Similarly, ∠ EHG = ∠ HGF = ∠ GFE = 90°

EFGH is a square with side “c”

Area of square = (a + b)²

Also, Area of square = Area of Constituent figures

= 2ab + c²

⇒ (a + b)² = 2ab + c²

⇒ a² + b² + 2ab = 2ab + c²

⇒ a² + b^{2 =} c²

This proves the Pythagoras Theorem, i.e. square of the hypotenuse is equal to the square of other two sides.