## Thursday, May 10, 2018

### Geometry Problem 1356: Quadrilateral, Triangle, Angle, 30 Degrees, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Let Angle B=4x, draw angle bisector BE of Angle B such that E lies on CD. Since ABC is isosceles Triangle, BE is also perpendicular bisector of AC. We get Angle BEC= Angle AEB = Angle AED = 60 Deg. Also Angle ABE = Angle CBE = 2x.

Consider Triangle ABE, ED is external bisector Angle AEB and ADB = Angle AEB/2 = 30 Deg. Hence D must be ex-center of Triangle ABE and BD must be bisector of Angle ABE
Hence Angle ABD = Angle EBD = x, We get Angle DBC = 3x = 3.Angle ABD .

1. To Agashe
https://photos.app.goo.gl/RKe0iP6Y79jH0Atf1
In my opinion, the statement of line 4 “Consider Triangle ABE, ED is external bisector Angle AEB and ADB = Angle AEB/2 = 30 Deg.” Is not enough to conclude that D is the ex-center of triangle ABE. Please provide more details . See above for the sketch .

2. If you draw a circle passing through AB such that the chord AB subtends angle AEB/2 at other points on circle. It will intersect CE at 2 ex-centers of Tr. ABE. According to diagram it must be ex-center opposite to B, hence D must be ex-center of Tr. ABE.

3. If O is the circumcenter of triangle ACD then triangle OAD is equilateral and O is the reflection of A in BD, thus <DBE=<ABD and BE is perpendicular bisector of AC, done.

4. To Agashe,

I still do not understand how you conclude that D is the ex-center of triangle ABE. Can you elaborate more, or post a diagram?

5. https://photos.app.goo.gl/qNBMmWbtA4FiGasX7

Stan has an excellent solution. See sketch in the above link for details
Peter Tran

2. According to previous coomments and the sketch of Peter
name P( BE meet circle), H (AP meet BD), L ( BD meet circle)
=> ang ALD = ang AED = 60°, ang LAP = LEP = 30° => AP perpendicular to BD
from conguence of right triangles BAH and BPH => BH bisector

3. Bisector angle ABC, intersect with CD at E,
Reflect triangle ABD along line BE, get triangle BCF; angle CBF =alpha; angle BFC = 30 degree;
Because angle AEB, angle BEC are 60 degree, so angle AED is 60 degree, so angle CEF is 60 degree;
So A,E,F on the same line;
Make equilateral triangle BCG, angle BEC = angle BGC = 60 degree;
So BCEG on the same circle;
Because angle GEB = angle AEB=60 degree,
So G, A,E,F on the same line;
So angle EGC = angle EBC = half angle ABC;
Also BG= CG, angle BFC=30 degree = ½ angle BGC;
So B,C,F on the same semicircle;
Angle FGC= 2x angle FBC = 2 alpha;
Angle FGC =angle EGC =half angle ABC =2 alpha;
So, alpha + theta = 2 x ( 2 alpha);
So theta= 3 alpha;

4. Reflect A in BD to create the point X. Then angle ADX is 60 degrees, and triangle ADX is equilateral. Hence angle AXD = 60 degrees.
Since angle ACD= 30 degrees, it must be on the circumference of the circle centred at X with radius XA. (Angle on circumference is half that at centre)
Therefore triangles AXB and CXB are congruent (SSS) so angle ABX = angle CBX
Hence 2alpha = theta - alpha and the required result follows immediately.

5. https://photos.app.goo.gl/cX2DnsWrXZg6DC4q9

Draw angle bisector BE of ∠ (ABC)
This bisector will perpendicular to AC and meet AC at E
So ∠ (CEB)=60=∠ (AEB)
Draw angle bisector EI of ∠ (BEA) => ∠ (BEI)= ∠ (AEI)= 30= ∠ (ADI)
Since ∠ (ADI)= ∠ (AEI)= 30 So ADEI is a cyclic quadri.
Since ∠ (ADI)= ∠ (FEI)=30 => central angle ∠ (AOI)= ∠ (IOF)=60
Triangle BAO congruent to BFO ( case SAS) => BD is the angle bisector of ∠ (ABE)
and ∠ (CBD)= 3. ∠ (ABD)

1. Peter: Can elaborate on this line: (FEI)=30 => central angle ∠ (AOI)= ∠ (IOF)=60 (Yes, ∠(FEI), but I do not see how that implies ∠ (IOF)=60 (F is not on the circle.)

2. ∠(IOF) is the central angle of ∠(IEF) since I, A, D,E,F is cyclic and O is the center
but ∠(IEF)=30 so ∠(IOF)=60

Peter Tran

6. Given ∠ACD = ∠ADB = 30°, and BA = BC.
Prove that ∠DBC = 3 times ∠ABD.

For E on CD, BE bisects ∠ABC. Connect AE. BE intersects AC at F. Then AE = CE. ∠BEC = ∠AEB = ∠AED = 60°. If ∠ABC = 4x, then ∠ABE =∠CBE = 2x. Bisect ∠BEA with EN, which meets BD at I. Connect AI; and because ∠IDA = 30° = 60° / 2 = ∠IEA, then IEDA is cyclic. Then ∠IED = 90° = ∠DAI.

Produce AI to meet BE at P. Consider quadrilateral PEDA:
∠APE = 360° -120° - 90° - ∠EDA = 150° - ∠EDA
= 60° + ∠DBE.

Consider Δ ABF: ∠BCF = ∠FAB
= 90° - 2x; ∠AID = 60°; ∠BIA = 120°; ∠IAB = 60° - ∠ABD.
∠FAI = 90° - ∠APF = 30° - ∠DBE. ∠EDB = 60° - ∠DBE.
∠PAB = 60° + ∠ABD - 2x = 60° - ∠DBE = ∠IAB = 60° - ∠ABD.
Thus ∠ABD = ∠DBE = x. And ∠DBC = 3x = ∠ABD * 3.

7. Take O the circumcenter of triangle ACD; ABCO is a kite, thus BO is angle bisector of <ABC. Triangle ADO is equilateral and O is reflection of A over BD, hence BD is bisector of <ABO, done.

1. Stan, can you elaborate on "O is reflection of A over BD, hence BD is bisector of <ABO"

8. See that O, circumcenter of triangle ADC is reflection of A in BD, which solves the problem.

9. Slight recasting of Stan’s solution with his kind permission

Let O be the circumcenter of Tr. ACD.
Then Tr. OAD is equilateral
ADOB is a kite & < DBO = alpha
ABCO is also a kite and so theta - alpha = 2.alpha & the result follows

10. Another Solution

Let BX be perpendicular to AC, X on AC
Draw a perpendicular from A to CD to meet BX extended at Y and CD at Z.

Since CXZY is concyclic, < AYB = 30 & so ABYD is concyclic

< CAB = 60 & Tr. AXZ is equilateral & since AY = CY, Tr. ACY is also equilateral
Hence AZ = ZY & so AD = DY.

Hence in concyclic ABYD, < DBY = alpha and so theta - alpha = 2.alpha which gives theta = 3.alpha

Sumith Peiris
Moratuwa
Sri Lanka