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.

GGeometry Problem 1356: Quadrilateral, Triangle, Angle, 30 Degrees, Congruence.

3 comments:

  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 .

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    Replies
    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 .

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    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.

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