Question

Write the converse of the Pythagoras Theorem and prove it.

Answer 1

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

Given: ${BC}^2={AB}^2+{AC}^2$ in $△ABC$

To prove that $\angle A$ is a right angle.

Let $\overline{OX}$ be any ray

we can construct $\overline{OY}$ such that $\overline{OY}\perp \overline{OX}$

Let $M\in OY$ such that $OM=AC$

Let $N\in \overline{OX}$ such that $ON=AB$

Draw $\overline{MN}$

$△OMN$ is right angle triangle as $\overline{OM}\perp \overline{ON}$
$\angle MON$ is a right angle
$\therefore$ $MN$ is the hypotenuse

According to the Pythagoras theorem
${MN}^2={OM}^2+{ON}^2={AC}^2+{AB}^2$
But ${AB}^2+{AC}^2={BC}^2$
${MN}^2={BC}^2$
$MN=BC......\left(i\right)$

$\therefore$ In $△ABC$ and $△ONM$ consider the correspondence $ABC\leftrightarrow ONM$
$\overline{AB}\cong \overline{ON}$ $(ON=AB)$
$\overline{AC}\cong \overline{OM}$ $(OM=AC)$
$\overline{BC}\cong \overline{MN}$ $(MN=BC)$

The correspondence $ABC\leftrightarrow ONM$ is congruence. So $△ABC\leftrightarrow △ONM$ (by SSS)
$\therefore$ $\angle A\cong \angle O$

But $\angle O$ in $△ONM$ is right angle by construction
$\therefore$ $\angle A$ is a right angle.