Answer :

Let the given right-angled triangle be ΔABC, right – angled at B.

We need to prove AC^{2} = AB^{2} + BC^{2}

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

i.e., BD⊥AC

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.

Hence, ΔADB~ΔABC

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

⇒ AD.AC = AB^{2}………(i)

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 = BC^{2}………(ii)

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

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

Now taking AC common on LHS, we get

⇒ AC (AD + CD) = BC^{2} + AB^{2}

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

⇒ AC(AC) = BC^{2} + AB^{2}

⇒ AC^{2} = BC^{2} + AB^{2}

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

Hence Proved.

Rate this question :