I had to use calculator to prove it

[Nadim Ul Abrar]

The triangle ABC has following property
* P is a interior point of triangle ABC such that
<PAB = 10 ( That means the angle PAB = 10 degree)
<PBA = 20
<PCA = 30
<PAC = 40
Prove that the triangle ABC is isosceles.

[Sakal]

Let,<PBC=x;<PCB=y...now, <APB=150,<APC=110 so, x+y=80 or,x=80-y;
In triangle ABP ,
AP/sin<ABP=AB/sin<APB
or, AP= ABsin<ABP/sin<APB
or, AP= ABsin 20/sin 150 =2ABsin 20...(1)
Similarly, from triangle APC we get,
AP =AC sin 30/sin 110=AC/(2cos 20)...(2)
Now ,dividing (1) by (2) we get ,
1=AB/AC. 4sin 20 cos 20
or, AC/AB= 2sin 40
or,sin<ABC/sin<ACB= 2sin 40 (from triangle ABC)
or, sin(20+x)/sin(30+y)=2sin40
or, sin(20+80-y)/sin(30+y)=2sin40
or, cos(10-y)=2sin40sin(30+y)
or,cos10cosy+sin10siny=sin40(cosy+3^(1/2)siny)
or,cosy/siny= (3^(1/2)sin40-sin10)/(cos10-sin40)
or, coty = (3^(1/2)(3^(1/2)/2.sin10+.5cos10)-sin10)/(cos10-(3^(1/2)/2.sin10+.5cos10))
or,coty= (1.5sin10+3^(1/2)/2cos10-sin10)/(cos10-.5cos10-3^(1/2)/2.sin10)
or,coty= (.5sin10+3^(1/2)/2cos10)/(.5cos10-3^(1/2)/2.sin10)
or,coty =cos(30-10)/sin(30-10)=cot20
so,y=20
Now,<BAC=40+10=50 and <ACB=30+y=30+20 =50 so.<BAC=<ACB hence,Triangle ABC is isosceles.