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(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{AQB}\)

(ii) BP = BQ or B is equidistant from the arms of \(\angle{A}\).

Answer :

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.

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