## Monday, February 25, 2019

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

Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Ajit Athle.
Details: Click on the figure below. 1. x=48
Nice problem

2. Draw CP=CB (P on AC) Join P to I, P to B, A to I.
APIB isosceles trapezoid => B/2 = 33°
=> x = 48°

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

How is AXIB an isosceles trapezoid?
AX = BI but that’s all?

2. Ang BAX + AXI = 180°

3. 4. Ang of ∆AXI 16,5° + 16,5° +147°
I was curious to know your solution for 1261

5. To c.t.e.o

Problem 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

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

4. Mark on CB extended BD = BI

Tr.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

1. Last line should be x = 48 not B = 48

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

Corollary is also true