- #1

- 16

- 0

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????? :rofl:thx

- Thread starter chickenguy
- Start date

- #1

- 16

- 0

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????? :rofl:thx

- #2

- 50

- 0

just draw and identify isosceles triangles.

You will get 40 degrees

You will get 40 degrees

- #3

- 16

- 0

i did what you said, but angle ABE was 70*

- #4

- 50

- 0

ABE=40 degress.

I think I have already posted it

I think I have already posted it

- #5

- 5

- 0

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

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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