ABC is an isosceles triangle with AB=AC and D is a point on ABC BC such that `AD_|_BC` (see figure ). To prove that `angleBAD =angleCAD ` a student proceeded as follows

`In Delta ABD and Delta ACD`,we have
` AB=AC " " [" Given"]`
`angleB=angleC " "[because` AB=AC ]
`and " "angleADB= angleADC `
Therefore` " " DeltaABD cong Delta ACD " " ["by AAS congruence rule"]`
So , ` " "angleBAD =angleCAD " " [by CPCT]`
What is the defect in the above argument ?

Explanation

In `Delta ABC ," " AB=AC " "` [given]
`rArr " " angleACB=angleABC " "` [angles opposite to equal sides are equal ]
in `Delta ABD and Delta ACD ,`

AB=BC
`angleABD = angleACD " " ["given "]`
`angleABD = angleACD " "["proved above "]`
`angleADB=angleADC " " ["each " 90^(@)]`
`therefore " "DeltaABD cong Delta ACD " "["by AAS "]`
So `" " angleBAD= angle CAD " " [" by CPCT "]`
So the defect in the given argument in that firstly prove `angleABD =angle ACD ` Hence ,` angleABD=anlge ACD ` is defect ,