Question

AB is a line segment C and D are points on opposite sides of AB such that each of them is equidistant from the point A and B as in the figure. The line CD is perpendicular bisector of AB.

• A. True
• B. False

Answer 1

The correct option is A True

$C$ is equidistant from points $A$ and $B$

$CA=CB$ and $D$ is equidistant from points $A$ and $B$

$⟹DA=DB$

We have to prove that $CD\perp AB$

$AD=BD$ and $\angle CPA=\angle CPB={90}^{\circ }$

In $△CAD$ and $△CBD$

$AC=BC$ ....... $\left(1\right)$

$AD=BD$ ........ $\left(2\right)$

$⟹CD=CD$ ......... (common)

$\therefore △CAD\cong △CBD$ ..... (SSS congruence rule)

Hence $\angle ACD=\angle BCD$ ........ (C.P.C.T) $\left(3\right)$

In $△CAP\cong △CBP$ ........ (SAS congruence rule)

$\therefore AP=BP(C.P.C.T)$

$⟹\angle APC=\angle BPC$ ........ (C.P.C.T)

Since $AB$ is a line segment

$\angle APC+\angle BPC={180}^{\circ }$

$⟹2\angle APC={180}^{\circ }\therefore \angle APC={90}^{\circ }$

$\angle APC=\angle BPC={90}^{\circ }$

Hence, $AC=BC$ and $\angle APC=\angle BPC={90}^{\circ }$

$\therefore CD$ is $\perp$ bisector of $AB$