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

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

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## Wednesday, December 19, 2018

### Geometry Problem 1410: Right Triangle, Incircle, Excircle, Tangency Points, 45 Degree Angle

Labels:
45 degrees,
angle,
excircle,
geometry problem,
incircle,
right triangle,
tangency point

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P intersection of diagonals of trapesoid. Diagonals are perpendicular

ReplyDeletetr FBD similar tr FPT (T tg point on AB)

=> FPG right isosceles triangle

https://photos.app.goo.gl/UNt6uwqEA6K9km9r6

ReplyDeleteper the result of problem 1409 we have F, D and M are colinear

We have ∠ (AIC)= 90+ B/2= 135 => ∠ (FIC)= 45

We have IC⊥MD and IE⊥ FG=> quadri. INPF is cyclic

And ∠ (DFP)= ∠ (EIN)= 45

Let X be the tangency point on AC of incircle

ReplyDeleteLet Y be the tangency point on BC of excircle

Considering the angles of quadrilateral CYFG, < GFY = 360 - (90-A/2) - (90 + A) - 135 = 45 - A/2 .....(1)

Now FDX is collinear as per my proof of Problem 1409 & so < XDC = 90 - C/2 = 45 + A/2.

Considering the angles of Tr. AFX, < BFD = 45 + A/2 - A = 45 - A/2 = < GFY from (1)

Since < BFD + < DFY = 45, < GFY + < DFY = Alpha = 45

Sumith Peiris

Moratuwa

Sri Lanka

Much simpler proof

ReplyDeleteConsidering Tr. XFG, < FXG = 90 - C/2 & < FGX = 90 - A/2, Alpha = A/2 + C/2 = 45

Sumith Peiris

Moratuwa

Sri Lanka

Considering usual triangle notations

ReplyDeleteLet U & V be the tangent points of incircle I with AB and AC

=> UV//FG and UF=VG=a => UVGF is an isosceles trepezoid-----(1)

It can be deduced that the diagonals of UVGF are congruent and say X be the point of intersection => GX=FX and FXG is isosceles ------(2)

Let W be the tangent point of excircle E with BC

It can be proved that W lies on the diagonal GU

(Apply menelaus to ABC => GC/GA*CW/WB*BU/UA =1

substituting and simplifying will give us the below

=> s(s-b)=(s-a)(s-c)

=> (a+b+c)(a+c-b)=(b+c-a)(a+b-c)

=> a^2+c^2=b^2

=> G,W&U are collinear

)

By symmetry, F,D&V are collinear -----------(3)

Extend BC to meet FG at Y, Since m(B)=90 and m(BFG)=90-A/2=> m(BYF)=A/2

Consider the triangle WYG, the external angle m(FGW)=m(YWG)+m(WYG)

=>m(FGW)=45-A/2+A/2

=>m(FGW)=45 ---------(4)

From (2) & (4) => m(XFG)=45 Q.E.D