Got it Connect C to I. Since I is the in centre angle ICB=C/2 Concentrate on that circle. Arc BI subtends angle ICB and Angle I,I(A),B. ~ ICB=I,I(A)B Since angle ILB=C ILB = angle LBI(A) +angle L,I(A)B {enterior angle} C=angle I(A)BL+C/2 Therefore angle I(A)BL=C/2
Couldn't you prove that B, I, C, and I_a are concyclic since IBI_a and ICI_a are both 90? Then you already proved that BL=IL=CL, so L must be the center of the circumcircle of BCI, but that circle also contains I_a. Then your done and the prove could have stopped at 8:10 . But still a very good proof in general!
7 yrs And Still Learning outcomes 🎉 thnx
Since angle A,I(A),B =C/2
Angle ACB=C
Can I say there exists a circle with centre at C and it passes through A,B and I(A)
No since that would assume that AC=BC=I_aC but that would assume ABC is iscoceles, but ABC is any regular triangle.
thank you very nice proof
thank you sir, this lemma helped me in solving a problem
how angle I,B,I(A)=90° explained here down in the comment
Got it
Connect C to I. Since I is the in centre angle ICB=C/2
Concentrate on that circle.
Arc BI subtends angle ICB and Angle I,I(A),B. ~ ICB=I,I(A)B
Since angle ILB=C
ILB = angle LBI(A) +angle L,I(A)B {enterior angle}
C=angle I(A)BL+C/2
Therefore angle I(A)BL=C/2
Thank you.
right angle good enough
Couldn't you prove that B, I, C, and I_a are concyclic since IBI_a and ICI_a are both 90? Then you already proved that BL=IL=CL, so L must be the center of the circumcircle of BCI, but that circle also contains I_a. Then your done and the prove could have stopped at 8:10 .
But still a very good proof in general!
Your proof is very good indeed. Thanks for sharing!
It is a mistake to state a right angle and then prove isosceles triangle.