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# 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.