Question

Write converse of Pythagoras theorem and prove it.

Answer 1

Converse of Pythagoras Theorem.

Statement:
In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Proof:

Here, we are given a triangle ABC in which $A{C}^2=A{B}^2+B{C}^2$.

We need to prove that $\angle B={90}^o$.

To start with, we construct a $\Delta$PQR right-angled at Q such that $PQ=AB$ and $QR=BC.$

Now, from $\Delta$ PQR, we have:

$P{R}^2=P{Q}^2+Q{R}^2$ (Pythagoras Theorem, as $\angle Q={90}^o$)

or $P{R}^2=A{B}^2+B{C}^2$ (By construction)........ (1)

But $A{C}^2=A{B}^2+B{C}^2$ (Given).......... (2)

So, $AC=PR$ (From (1) and (2)).............. (3)

Now, in $\Delta ABC$ and $\Delta PQR$,

$AB=PQ$ (By construction)

$BC=QR$ (By construction)

$AC=PR$ (Proved in (3))

So, $\Delta ABC\simeq \Delta PQR$ (SSS congruence)

$\angle B=\angle Q$ (Corresponding angles of congruent triangles)

$\angle Q={90}^o$ (By construction)

So $\angle B={90}^o$