In the triangle $ABC$, the bisector of $\angle A$ intersects the bisection of $\angle B$ at the point $I, D$ is the foot of the perpendicular from $I$ onto $BC$. Prove that the bisector of $\angle BIC$ is perpendicular to the bisector $\angle AID$.
Problem
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Tags: geometry, incenter
NikoIsLife
10.07.2019 15:10
parmenides51 wrote: Prove that the bisector of $\angle BIC$ is perpendicular to $\angle AID$. Sorry, but what does "perpendicular to $\angle AID$" mean?
parmenides51
10.07.2019 15:17
thanks, corrected
Chotu2004
10.07.2019 15:44
can be simply done by angle chasing. the angle between these two can be simply obtained as 90 -45 + ( a+b+c)/4 90-45+45=90°
Kagebaka
10.07.2019 17:14
Notice that it suffices to show that $ID$ and $AI$ are isogonal in $\angle BIC.$ However, this is obvious since $AI$ passes through the circumcenter of $\triangle BIC$ by the Incenter-Excenter Lemma and $ID$ is the $I$-altitude. $\blacksquare$
parmenides51
24.05.2024 17:38
A solution that is solved only by angles triangle sum is strange to find at this level
I is the incenter so CI bisects <ACB
Let <BAI=<IAC =x, < AIB=<IBC = y , <ACI=<ICB = z
By angles triangle sum in ABC : 2x+2y+2z= 180^o <=> x+ y + z= 90^o (1)
Let bisector of <AIB intersect AB at K, let the bisector of <BIC intersect BL at L
Let <AKI = I_1, <KIB = I_2, <BID = I_3, <DIZ = I_4, <ZIC = I_5
From angle bisector we get I_1 = I_2+ I_3 (2) and I_5= I_3+I_4
We want to prove that <KiD = 90^o <=>
I_2+ I_3 + I_4= 90^o <=>
2I_2+2 I_3 +2 I_4= 180^o <=>
I_2+ (I_2+ I_3 ) +(I_3 + I_4) + I_4= 180^o <=> (because of 2 and 3)
I_2+ I_1+ I_5 + I_4= 180^o <=>
(I_ 2+ I_1+)+(I_5 + I_4) =180^o <=>
90^o + x + 90^o - x = 180^o, which is true