Iftwoangles of a triangleare equal;then sides opposite to them are also equal.Ifthe altitude from one vertex of atriangle bisectsthe opposite side; then thetriangle is isosceles. ...If the bisectorof thevertical angle of a triangle bisects the baseof thetriangle;then thetriangle is isosceles.

Given:In ∆ABC , AD bisects ∠BAC, & BD= CD

To Prove:AB=AC

Construction:Produce AD to E such that AD=DE & then join E to C.

Proof:

In ∆ADB & ∆EDCAD= ED ( by construction)∠ADB= ∠EDC. (vertically opposite angles (

BD= CD (given)

∆ADB congruent ∆EDC (by SAS)

Hence, ∠BAD=∠CED......(1) (CPCT)

∠BAD=∠CAD......(2). (given)

From eq.1 &2 ∠CED =∠CAD......(3)

AB=CE (CPCT).......(4)

From eq 3 as proved that

∠CED=∠CAD

So we can say CA=CE......(5)

[SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL]

Hence, from eq 4 & 5

AB = AC

HENCE THE ∆ IS ISOSCELES..

Given:In ∆ABC , AD bisects ∠BAC, & BD= CDTo Prove:AB=ACConstruction:Produce AD to E such that AD=DE & then join E to C.Proof:In ∆ADB & ∆EDCAD= ED ( by construction)∠ADB= ∠EDC. (vertically opposite angles (BD= CD (given)∆ADB congruent ∆EDC (by SAS)Hence, ∠BAD=∠CED......(1) (CPCT)∠BAD=∠CAD......(2). (given)From eq.1 &2 ∠CED =∠CAD......(3) AB=CE (CPCT).......(4)From eq 3 as proved that ∠CED=∠CADSo we can say CA=CE......(5)

[SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL]Hence, from eq 4 & 5 AB = ACHENCE THE ∆ IS ISOSCELES..