Incenter Excenter Lemma - Proof

From Evan Chen notes here: www.mit.edu/~ev...

[@Cooososoo]

7 yrs And Still Learning outcomes 🎉 thnx

[@akashsudhanshu5420]

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)

[@sallyxu4668]

No since that would assume that AC=BC=I_aC but that would assume ABC is iscoceles, but ABC is any regular triangle.

[@akashsudhanshu5420]

how angle I,B,I(A)=90° explained here down in the comment

[@akashsudhanshu5420]

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

[@rafeef039]

thank you very nice proof

[@sallyxu4668]

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!

[@OsmanNal]

Your proof is very good indeed. Thanks for sharing!

[@mauz791]

Thank you.

[@lewischeung5522]

right angle good enough

[@adityakiranbal9919]

thank you sir, this lemma helped me in solving a problem

[@lewischeung5522]

It is a mistake to state a right angle and then prove isosceles triangle.