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

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

This entry contributed by Ajit Athle.

Details: Click on the figure below.

## Monday, February 25, 2019

### Geometry Problem 1418: Triangle, Circumcircle, Incenter, Side, Sum of Segments, Tangent, Parallel, Angle

Labels:
angle,
angle bisector,
circumcircle,
geometry problem,
incenter,
parallel,
side,
sum of segments,
tangent

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x=48

ReplyDeleteNice problem

Draw CP=CB (P on AC) Join P to I, P to B, A to I.

ReplyDeleteAPIB isosceles trapezoid => B/2 = 33°

=> x = 48°

Better replace P by X (say) since P is already given

DeleteHow is AXIB an isosceles trapezoid?

AX = BI but that’s all?

Ang BAX + AXI = 180°

Deleteyes thanx

DeleteAng of ∆AXI 16,5° + 16,5° +147°

DeleteI was curious to know your solution for 1261

To c.t.e.o

DeleteProblem 1261

Mark X on AB such that BX = BC

Mark Y on AD such that DC = DY

Use Pythagoras and show CX = AY and AX = CY that is AXCD is a parallelogram.

X mid point of AB so C midpoint of BE

https://photos.app.goo.gl/kMimcwDGK9WHYCVs9

ReplyDeleteDefine points D and E per sketch

BD=BI so CA= CD

Triangles ACD and IBD are isosceles and CD⊥AD

Due to symmetry we have ∠ (IDC)= ∠ (IAC)=33/2=16.5

So ∠ (IBC)=33 and ∠ (ACP)=66

Note that ACP is isosceles and APCQ is cyclic

From above we calculate x=48

Mark on CB extended BD = BI

ReplyDeleteTr.s AIC & DIC are congruent SAS

So < IDC = < B/4 = < CAI 33/2 whence B = 66

Now < AQP = B = < PAC = < PCA = < PQC since APCQ is concyclic

So 2B + x = 180 and B = 48

Sumith Peiris

Moratuwa

Sri Lanka

Last line should be x = 48 not B = 48

DeleteSo we conclude that if in a Tr. ABC, BI + BC = AC where I is the incentre, then B = 2A

ReplyDeleteCorollary is also true