Question

Two lines are intersected by a transversal, prove that the bisector of interior angles form a right angle.

Answer 1

The interior angles on the same side of the transversal (called consecutive interior angles) are supplementary (they sum to 180 degrees). Call these angles a and b.

$a+b={180}^o$

Now divide both sides by $2$:

$\frac{a}{2}+\frac{b}{2}={90}^o$

$\frac{a}{2}$ and $\frac{b}{2}$ are just the halves of $a$ and $b$ formed by their bisectors.

These bisectors intersect, forming the legs of a triangle with the transversal being the third side.

The interior angles of this triangle are $\frac{a}{2},$ $\frac{b}{2}$ and the angle formed by the intersection of the bisectors.

Call this as $\angle c$.

Since the sum of the interior angles of a triangle is 180 degrees:

$\frac{a}{2}+\frac{b}{2}+c={180}^o$

Since you know $\frac{a}{2}+\frac{b}{2}={90}^o$

${90}^o+\angle c={180}^o$

$\Rightarrow \angle c={90}^o$

So the bisectors are perpendicular.