Question

Perpendicular bisectors of two sides of the triangle intersect each other at a point on the third side. Then show that the triangle is right angled triangle. (3.74)

Answer 1

$$\begin{gathered} \mathrm{Construction}:\mathrm{Draw}\mathrm{a}\mathrm{right}-\mathrm{angled}△\mathrm{ABC}\mathrm{and}\mathrm{name}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{AC},\mathrm{P}. \\ \mathrm{Join}\mathrm{PB}. \\ \mathrm{Draw}\mathrm{a}\mathrm{circle}\mathrm{circumscribing}\mathrm{the}\mathrm{triangle}. \\ \mathrm{By}\mathrm{perpendicula}-\mathrm{bisector}\mathrm{theorem}, \\ \mathrm{PA}=\mathrm{PB}\mathrm{and}\mathrm{PB}=\mathrm{PC} \\ \text{Or},\mathrm{PA}=\mathrm{PB}=\mathrm{PC} \\ \mathrm{In}\mathrm{the}\mathrm{circle},\mathrm{with}\mathrm{PA}=\mathrm{PB}=\mathrm{PC}\mathrm{as}\mathrm{radi}i,\text{P}\mathrm{becomes}\mathrm{the}\mathrm{centre}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}\mathrm{is}90°. \\ \text{T}\mathrm{hus},\angle \mathrm{ABC}=90°. \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$