ABC and DBC are two triangle on the same base BC such that A and D lie on the opposite sides of BC,AB=AC and DB =DC ,Show that AD is the perpedicular bisector of BC.

Explanation

Given two `Delta ABC and Delta DBC ` are formed on the same base BC such that A and D lie on the opposite sides of BC such that AB= AC and DB=DC .Also AD intersects BC at O.
To show AD is the Perpendicular bisector ot BC ie. `AD_|_BC ` and AD bisects BC .
Proof in `Delta ABD and Delta ACD `
AB=AC [given ]
AD=AD [common side ]
and BD=DC [given ]
`therefore DeltaABD cong Delta ACD ` [by SSS congruence rule ]
`rArr angle BAD = angle CAD `[by CPCT ]
in `Delta AOB and Delta AOC`

AB=AC [given]
AO =OA [common side ]
and ` angle BAO= angle CAO` [proved above ]
`therefore angle AOB cong Delta AOC ` [by CPCT]
but ` angle AOB + angle AOC =180^(@)` [linear pair axiom ]
`rArr angle AOB + angle AOB =180^(@)` [from Eq.(i) ] ltbr gt `rArr 2 angle AOB =180 ^(@)`
`rArr angle AOB =(180^(@))/(2) =90^(@)`
Hence `,AD _|_ BC `and AD bisects BC i.e AD is the perpendicualr bisector