Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Sunday, January 6, 2019

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

Subscribe to:
Post Comments (Atom)

Full explanation

DeleteT, 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

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

ReplyDeletePer 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

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

ReplyDeleteHence < 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

Considering usual triangle notations,

ReplyDeletem(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