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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.