## Sunday, January 6, 2019

### Geometry Problem 1411: Right Triangle, Incircle, Excircle, Tangency Points, Isosceles Right Triangle

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Full explanation
T, P, tg points on AB, AC
PTHG isosceles trapezoid => ang GPH = ang HTG => Tr GJH isosceles
Tr BTF ~ Tr DJF (TFB common ang, ang GTH = GPH = PDC) => GJH = 90

1. https://photos.app.goo.gl/UpEAXe6BoaArDLy39

Per the result of problem 1409 we have G, D, M are collinear
We have ∠ (AIC)= 90+ ∠ (B)/2 = 135 => ∠ (EIC)=45
We have GH⊥IE and DM⊥IC => ILKG is cyclic
So ∠ (LGK)= ∠ (KIC)=45
We have triangles FCH and MCD are isosceles
External angle ∠ (MCD)= 2 x ∠ (CHF)= 2.u
So ∠ (LCF)= ∠ (CFH)=u => CL//HJ => HJ⊥GJ
So GJH is an isosceles right triangle

2. If the tangency point of AC is X,GDJX are collinear points

Hence < AGX = 90 - C/2 - A = 45 - A/2 and since < BGF = 45, < JGF = A/2
But < FGG = C/2 so < JGH =A/2 + C/2 = 45

Now < AGH = 90 - A/2 and since < FHC = C/2, < JHG = 90 - A/2 - C/2 = 45

So in Tr. JGH, 2 angles are 45 each, hence it is right isosceles

Sumith Peiris
Moratuwa
Sri Lanka

3. Considering usual triangle notations,
m(CFH)=m(BFJ)=45-A/2 --------(1)
A,I,J & E are collinear => m(BEJ)=45-A/2 -------(2)
From (1)&(2) B,J,F&E are concyclic ---------(3)
Since m(BFE)=m(BGE)=90 => G also lies on the same circle as (3)
=> m(GJF)=m(GBF)=90 Q.E.D

4. In △ABC, ∠ACB = 90°, CD = median, in the line AC we assign point M, MC=MD.
O and N are circle centers outside the triangles △AMB and △CDM.
We know that AB=C, find ON.

Erina NJ