Given in `Delta ABC ` D is the mid point of AC i.e AD=CD such that BD `=1/2` AC
TO show `angle ABC =90 ^(@)`
Proof We have ` BD=(1)/(2) AC `
Since D is the-point of AC
`AD=CD =(1)/(2)AC ...(ii) `
From Eqs. (i) and (ii)
AD=CD=BD
In `Delta DAB,` AD=BD [Proved above ]
`therefore angleABD = angle BAD ...(iii)`
[angles opposite to equal sides are euql ]
In `Delta DBC ` BD=CD [Proved above ]
`therefore angleBCD=angle CBD ....(iv)`
[angles opposite to equal sides are equal ]
In `Delta ABC " "angle ABC + angle BAC + angle ACB =180^(@)`
[by angle sum property of a triangle ]
`rArr " "angleABC+angleBAD+angleDCB=180^(@)`
`rArr " " angleABC+angleABD+angleCBD=180^(@)`
`rArr " "angle ABC+angle ABC=180^(@)`
`rArr " "2angleABC=180^(@)`
`rArr " " angle ABC=90^(@)`
