Answer :

Given: ∆ABC is right angled triangle at B, where AC = hypotenuse, AB = base and BC = perpendicular

To prove: (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}

⇒ (AC)^{2} = (BC)^{2} + (AB)^{2}

Construction: Draw BD perpendicular to AC.

Proof: Since BD is perpendicular to AC,

A theorem is applicable here, which says that if a perpendicular is drawn from the vertex of a right angle of a right triangle to the hypotenuse, then the triangles formed on either sides of the perpendicular are similar to the whole triangle as well as to each other.

This theorem simply means that, ∆ADB ∼ ∆ABC and ∆BDC ∼ ∆ABC as well as ∆ADB ∼ ∆BDC.

Thus, if ∆ADB ∼ ∆ABC, by similar triangle property we can write,

⇒

⇒ AD.AC = AB^{2} …(i)

Similarly, if ∆BDC ∼ ∆ABC

⇒

⇒ DC.AC = BC^{2} …(ii)

Now we need to add equations (i) and (ii), we get

AD.AC + DC.AC = AB^{2} + BC^{2}

⇒ AC × (AD + DC) = AB^{2} + BC^{2}[∵ AC is common in AD.AC + DC.AC, AC is taken common]

⇒ AC × AC = AB^{2} + BC^{2} [∵ AD + DC is clearly AC from the diagram]

⇒ AC^{2} = AB^{2} + BC^{2}

Hence, it is proved.

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