What is the size of angle ABE?

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Discussion Overview

The discussion revolves around a geometry problem involving a triangle ABC and the determination of the size of angle ABE based on given conditions and relationships between various segments and angles. The scope includes mathematical reasoning and problem-solving related to triangle properties and angle relationships.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests drawing and identifying isosceles triangles to find angle ABE, proposing it to be 40 degrees.
  • Another participant claims to have calculated angle ABE as 70 degrees after following the previous suggestion.
  • A later post provides a detailed solution, arriving at angle ABE being 40 degrees through a series of angle relationships and properties of triangles.
  • There is a mention of the relationships AD=BC and BE=DC as given conditions that influence the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the size of angle ABE, with one participant asserting it is 70 degrees and others claiming it is 40 degrees. The discussion remains unresolved regarding the correct size of angle ABE.

Contextual Notes

The discussion includes assumptions about the properties of isosceles triangles and angle relationships, but there are no explicit confirmations of these assumptions or the correctness of the calculations presented.

chickenguy
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hi everyone, there si this problem that i can't figure out and i was looking for some help
FIND THAT ANGLE!

In a triangle ABC, angleBAC=35*, D is a point on AB such that AD=BC and angle ADC=110*. E is a point on AC such that BE=DC. What is the size of angle ABE? :smile:thx
 
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just draw and identify isosceles triangles.
You will get 40 degrees
 
i did what you said, but angle ABE was 70*
 
ABE=40 degress.
I think I have already posted it
 
The solution:
The symbol of degree is omitted.

AD=BC (given)
BE=DC (given)
angleACD=180-angleBAC-angleADC (angle sum of triangle)
=180-35-110
=35
Because angleACD=angleBAC
Therefore DA=DC (sides opp. eq. angle)
angleCDB=angleACD+angleCAD (ext. angle of triangle)
=35+35
=70
DA=DC
DC=BC
angleCBD=angleCDB (base angles, isos. triangle)
=70
angleDCB=180-angleCBD-angleCDB (angle sum of triangle)
=180-70-70
=40
DC=BC
BE=BC
angleCEB=angleECB (base angles, isos. triangle)
=angleACD+angleDCB
=35+40
=75
angleABE+angleBAC=angleCEB (ext. angle of triangle)
angleABE+35=75
angleABE=40
 
Last edited:

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