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Shenghui Yang

[Notebook] Excenters and Orthic Triangle

Shenghui Yang, WOLFRAM

Posted 4 months ago

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Problem Let ABC be a triangle with I A , I B , and I C as excenters. Prove that excentral triangle I A I B I C has orthocenter I H and that triangle ABC is its orthic triangle ( pedal triangle centered on orthocenter ). Geometric Scene Construction Though we can use GeometricScene to generate arbitrary triangle, we can specify some coordinates for our demo triangle to make the diagram with visually pleasant ratio. This code is rather simple with repetitive TriangleCenter calls. gs=GeometricScene[{a{0,0},b{4,0},c{3.1,3.5},ia,ib,ic},{ia==TriangleCenter[{a,b,c},{"Excenter",a}],ib==TriangleCenter[{a,b,c},{"Excenter",b}],ic==TriangleCenter[{a,b,c},{"Excenter",c}],TriangleConstruct[{ia,ib,ic},"Orthocenter"]}]//RandomInstance;Framed[gs] Out[]= Proof by Angle Chasing Colinear property. WLOG, you only need I B C I A sit on the same line. Notice that ∠I B CA= ∠I A CB . We can quickly check ◼ ∠I B CA is 1 2 (π-∠BCA) PlanarAngle[{ib,c,a}]/.gs["Points"] 1.08252 Out[]= and ∠I A CB is 1 2 (π-∠BCA) PlanarAngle[{ia,c,b}]/.gs["Points"] 1.08252 Out[]= which is very clear ∠I A CI B = ∠I B CA+∠BCA+ ∠I A CB=π . Done. Perpendicular property. WLOG, we only need line CI c is perpendicular to I A I B . This is simple because we know I C is excenter, it implies that CI c is angle bisector of ∠BCA . ∠I B CI C = ∠I B CA+ ∠I C CA= ∠I C CB+ ∠I A CB= ∠I B CI A . We also know ∠I B CI C + ∠I C CI A = ∠I A CI B =π . Thus, ∠I B CI C =π/2 . Done. ◼ PlanarAngle[{ib,c,a}]+PlanarAngle[{ic,c,a}]/.gs["Points"] 1.5708 Out[]= %/Degree 90. Out[]= Therefore we have BI B ⊥ I C I A and AI A ⊥ I C I B . This implies triangle ABC is the orthic triangle of triangle I A I B I C . Additionally, we have some similarity property. WLOG, we only have to prove CAI B ∼ BCI A . We need one more angle equation besides ∠I B CA= ∠I A CB . Notice that ◼ ∠I A BC= 1 2 (π-∠ABC)and ∠AI B C= 1 2 (π- ∠I B AC- ∠I B CA)= 1 2 π- 1 2 (π-∠BAC)- 1 2 (π-∠BCA)= 1 2 (∠BAC+∠ACB)= 1 2 (π-∠ABC)= ∠I A BC Thus CAI B ∼ BCI A . We can prove other similar triangle pairs in the same way. Use FindGeometricConjectures to list them FindGeometricConjectures[gs,_[__,"Similar"]]["Conclusions"] {GeometricAssertion[{Triangle[{ia,b,c}],Triangle[{a,ib,c}],Triangle[{a,b,ic}],Triangle[{ia,ib,ic}]},Similar]} Out[]= Review Use excenter’s definition and angle bisectors ◼ Arithmetic with neighboring angles to find perpendicular segments ◼ Find similar triangles with angular addition and interior angles in a triangle sum up to π ◼ Other Geometry Problems Summable Areas ◼ Circumcenter, Midpoint and Pi ◼

POSTED BY: Shenghui Yang

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[Notebook] Excenters and Orthic Triangle

Problem

Geometric Scene Construction

Proof by Angle Chasing

Review

Other Geometry Problems