Question

As shown in the fig. 3.14 a semi circle with centre O and diameter AB is drawn. Keeping the radius constant, arcs are drawn with points A and B as centres. These arcs intersect the semi circle at points P and Q respectively. Similarly two more arcs are drawn. keeping radius constant, with centres P and Q. These arcs intersect at point X. Show that seg XO $\perp$ line AB.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{AB}\mathrm{is}\mathrm{the}\mathrm{diameter}. \\ \mathrm{All}\mathrm{arcs}\mathrm{are}\mathrm{drawn}\mathrm{with}\mathrm{the}\mathrm{same}\mathrm{radius}. \\ \mathrm{Construction}:\mathrm{Join}\mathrm{PX},\mathrm{QX},\mathrm{PA},\mathrm{QB},\mathrm{PO}\mathrm{and}\mathrm{QO}. \\ \mathrm{In}△\mathrm{AOP}\mathrm{and}△\mathrm{BOQ}, \\ \mathrm{AO}=\mathrm{OB}(\because \mathrm{O}\mathrm{is}\mathrm{the}\mathrm{midpoint}) \\ \mathrm{AP}=\mathrm{QB}(\mathrm{arcs}\mathrm{are}\mathrm{cut}\mathrm{with}\mathrm{same}\mathrm{radius}) \\ \mathrm{PO}=\mathrm{QO}\left(\text{r}\mathrm{adius}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{circle}\right) \\ \therefore △\mathrm{AOP}\cong △\mathrm{BOQ}(\mathrm{SSS}\mathrm{test}) \\ \mathrm{Thus},\angle \mathrm{AOP}=\angle \mathrm{BOQ}(\mathrm{by}\mathrm{c}.\mathrm{a}.\mathrm{c}.\mathrm{t})....\left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{POX}\mathrm{and}△\mathrm{QOX}, \\ \mathrm{PO}=\mathrm{QO}\left(\text{r}\mathrm{adius}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{circle}\right) \\ \mathrm{PX}=\mathrm{QX}(\mathrm{arcs}\mathrm{are}\mathrm{cut}\mathrm{with}\mathrm{same}\mathrm{radius}) \\ \mathrm{OX}=\mathrm{OX}\left(\text{c}\mathrm{ommon}\right) \\ \therefore △\mathrm{POX}\cong △\mathrm{QOX}(\mathrm{SSS}\mathrm{test}) \\ \mathrm{Thus},\angle \mathrm{POX}=\angle \mathrm{QOX}(\mathrm{by}\mathrm{c}.\mathrm{a}.\mathrm{c}.\mathrm{t})......\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{AOX}=\angle \mathrm{BOX} \end{gathered}$$

$$\begin{gathered} \mathrm{Since},\angle \mathrm{AOX}+\angle \mathrm{BOX}=180°\left(\text{l}\mathrm{inear}\mathrm{pair}\right) \\ \Rightarrow 2\angle \mathrm{AOX}=180° \\ \Rightarrow \angle \mathrm{AOX}=90°=\angle \mathrm{BOX} \end{gathered}$$

Hence, seg XO is perpendicular to line AB.