Answer :

Given: In Δ ABC, AB which intersects AB and AC at D and F respectively.

To Prove:

Construction:

Join B, E and C, D and then draw DM ⟘ AC and EN ⟘ AB.

Proof:

Area of Δ ADE

Area of Δ BDE

So

… (2)

Again, area of Δ ADE

Area of Δ CDE

So

…(3)

Observe that Δ BDE and Δ CDE are on the same base DE and between same parallels BC and DE.

So, ar(Δ BDE) = ar(Δ CDE) … (3)

From (1), (2) and (3), we have

Hence, proved.

**OR**

Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

Given: A right triangle ABC right angled at B.

To Prove: AC^{2} = AB^{2} + BC^{2}

Construction: Draw BD ⟘ AC

Proof:

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

⇒ Δ ADB ~ Δ ABC

So,

Or AD × AC = AB^{2} … (1)

Also, Δ BDC ~ Δ ABC

So,

Or CD × AC = BC^{2} … (2)

Adding (1) and (2),

AD × AC + CD × AC = AB^{2} + BC^{2}

⇒ AC (AD + CD) = AB^{2} + BC^{2}

⇒ AC^{2} = AB^{2} + BC^{2}

Hence, proved.

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