Line l is the bisector of a \(\angle{A}\) and \(\angle{B}\) is any point on l. BP and BQ are perpendiculars from B to the arms of \(\angle{A}\) (see figure). show that
(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\)
(ii) BP = BQ or B is equidistant from the arms of \(\angle{A}\).

In \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\), we have

\(\angle{APB}\) = \(\angle{AQB}\) = \(90^\circ\)
\(\angle{PAB}\) = \(\angle{QAB}\) ...(AB bisects \(\angle{PAQ}\))

AB is the common side.
\(\therefore \) \(\triangle \) {APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\) ...(By AAS congruency test)

Also, BP = BQ ...(By CPCT)

Thus, we can say that, B is equidistant from the arms of \(\angle{A}\).
Hence, proved.