# Prove that in a r

Let the given right - angled triangle be ΔABC, right - angled at B. We need to prove AC2 = AB2 + BC2

Now let’s draw a perpendicular line BD to line AC

i.e., BDAC

Now we know,

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.

Now applying the similar triangles condition, i.e., sides of similar triangles are same in ratio, we get Now similarly consider the other set of similar triangles,

ΔBDC~ΔABC

Now applying the similar triangles condition, i.e., sides of similar triangles are same in ratio, we get CD.AC = BC2………(ii)

Now adding equation (i) and (ii), we get

AD.AC + CD.AC = BC2 + AB2

Now taking AC common on LHS, we get

AC (AD + CD) = BC2 + AB2

Now from figure we can see that AD + CD = AC, so above equation becomes

AC(AC) = BC2 + AB2

AC2 = BC2 + AB2

Hence in a right - angled triangle square of the hypotenuse is equal to sum of the squares of the other two sides.

Hence Proved.

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