Question

AB is a line segment. C and D are points on opposite sides of AB such that each of them is eqidistace from A and B. Show that the line CD is the perpendicular bisector of AB.​

Answer 1

Answer

C is equidistant from points A and B

CA=CB and D is equidistant from points A andB

⟹DA=DB

We have to prove that CD⊥AB

AD=BD and ∠CPA=∠CPB=90∘

In △CAD and △CBD

AC=BC ....... (1)

AD=BD ........ (2)

⟹CD=CD ......... (common)

∴△CAD≅△CBD ..... (SSS congruence rule)

Hence ∠ACD=∠BCD ........ (C.P.C.T) (3)

In △CAP≅△CBP ........ (SAS congruence rule)

∴AP=BP(C.P.C.T)

⟹∠APC=∠BPC ........ (C.P.C.T)

Since AB is a line segment

∠APC+∠BPC=180∘

⟹2∠APC=180∘

∴∠APC=90∘

∠APC=∠BPC=90∘

Hence, AC=BC and ∠APC=∠BPC=90∘

∴CD is ⊥ bisector of AB