Answer :

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]

BD

So,

Using Pythagoras Theorem,

**(Hypotenuse)**

^{2}= (Base)^{2}+ (Perpendicular)^{2}AE^{2 }= AC^{2 }– EC^{2}

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

And, AB^{2} = x^{2}

∴

**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

**SAS rule:**Side-Angle-Side is a rule used to prove whether a given set of triangles are congruent. If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.

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 + c^{2}

⇒ (a + b)^{2} = 2ab + c^{2}

⇒ a^{2} + b^{2} + 2ab = 2ab + c^{2}

⇒ a^{2} + b^{2 }= c^{2}

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

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