Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Thursday, May 10, 2018
Geometry Problem 1356: Quadrilateral, Triangle, Angle, 30 Degrees, Congruence
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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.
ReplyDeleteConsider 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 .
To Agashe
Deletehttps://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 .
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.
DeleteIf 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.
DeleteTo Agashe,
DeleteI still do not understand how you conclude that D is the ex-center of triangle ABE. Can you elaborate more, or post a diagram?
https://photos.app.goo.gl/qNBMmWbtA4FiGasX7
DeleteStan has an excellent solution. See sketch in the above link for details
Peter Tran
According to previous coomments and the sketch of Peter
ReplyDeletename 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
Bisector angle ABC, intersect with CD at E,
ReplyDeleteReflect 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;
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.
ReplyDeleteSince 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.
https://photos.app.goo.gl/cX2DnsWrXZg6DC4q9
ReplyDeleteDraw 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)
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.)
Delete∠(IOF) is the central angle of ∠(IEF) since I, A, D,E,F is cyclic and O is the center
Deletebut ∠(IEF)=30 so ∠(IOF)=60
Peter Tran
Given ∠ACD = ∠ADB = 30°, and BA = BC.
ReplyDeleteProve 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.
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.
ReplyDeleteStan, can you elaborate on "O is reflection of A over BD, hence BD is bisector of <ABO"
DeleteSee that O, circumcenter of triangle ADC is reflection of A in BD, which solves the problem.
ReplyDeleteSlight recasting of Stan’s solution with his kind permission
ReplyDeleteLet 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
Another Solution
ReplyDeleteLet 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