Question

Write the converse of Pythagoras theorem and prove it.

Answer 1

Define the converse of Pythagoras' theorem and prove the same:

The converse of Pythagoras' theorem states that if the square of one side of a triangle equals the sum of the squares of the other two sides, the angle opposite the longest side equals $90$ degrees.

Consider the following two triangles:

In triangle $ABC,A{C}^2=A{B}^2+B{C}^2$

We need to prove that $\angle B=90°$.

Construct another triangle $PQR$ such that $PQ=AB,QR=BC,\angle Q=90°$.

The square of the hypotenuse side in a right-angled triangle is equal to the sum of the squares of the other two sides, according to the Pythagoras theorem.

So,

$P{R}^2=P{Q}^2+Q{R}^2...\left(1\right)$

Put

$PQ=AB,QR=BC$ in $\left(1\right)$, we get

$\begin{array}{rcl}P{R}^2& =& A{B}^2+B{C}^2\end{array}$

Also,

$A{C}^2=A{B}^2+B{C}^2$.

Therefore,

$P{R}^2=A{C}^2$ that is $PR=AC$.

In triangles $ABC,PQR$,

$\begin{array}{rcl}AB& =& PQ\\ BC& =& QR\\ AC& =& PR\end{array}$

So, $∆ABC\cong ∆PQR$ (by SSS congruency criteria )

$\begin{array}{rcl}\angle B& =& \angle Q\\ \therefore \angle B& =& 90°\left[\because \angle Q=90°\right]\end{array}$

Hence, the converse of Pythagoras' theorem states that if the square of one side of a triangle equals the sum of the squares of the other two sides, the angle opposite the longest side equals 90 degrees also, the theorem has been proved.