If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.
Answer :
We are given a point D on side BC of a ABC such that
BAD = CAD and BD = CD
We want to prove that AB =AC.
Now we produce our line AD to E , As AD = DE , Then join CE , Now we have our figure , As :
![](https://s3mn.mnimgs.com/img/shared/content_ck_images/images/fig12-9-6(1).jpg)
Now In ABD and ECD
BD = CD ( Given )
AD = ED ( By construction )
ADB = EDC ( Vertically opposite angles )
Hence
ABD ECD ( By SAS rule )
So,
AB = EC -------------------- ( 1 ) ( BY CPCT )
And
BAD = CED -------------------- ( 2 ) ( BY CPCT )
So we get
CED = CAD ( As given BAD = CAD )
So from base angle theorem we get
AC = EC
So,
AB = AC ( As we know from equation 1 , AB = EC )
Hence
In ABC , AB = AC , So ABC is a isosceles triangle . ( Hence proved )
We are given a point D on side BC of a ABC such that
BAD = CAD and BD = CD
We want to prove that AB =AC.
Now we produce our line AD to E , As AD = DE , Then join CE , Now we have our figure , As :
![](https://s3mn.mnimgs.com/img/shared/content_ck_images/images/fig12-9-6(1).jpg)
Now In ABD and ECD
BD = CD ( Given )
AD = ED ( By construction )
ADB = EDC ( Vertically opposite angles )
Hence
ABD ECD ( By SAS rule )
So,
AB = EC -------------------- ( 1 ) ( BY CPCT )
And
BAD = CED -------------------- ( 2 ) ( BY CPCT )
So we get
CED = CAD ( As given BAD = CAD )
So from base angle theorem we get
AC = EC
So,
AB = AC ( As we know from equation 1 , AB = EC )
Hence
In ABC , AB = AC , So ABC is a isosceles triangle . ( Hence proved )