Equilateral triangle ABC and isosceles triangle DBC share side BC. If angle BDC= 34 and BD=BC, what is the measure of angle ABD?
I don't know how to solve this questions, can someone guide me through?
See other post.
To solve this question, we need to use the properties of angles in triangles. Here's a step-by-step guide on how to solve it:
1. Start by drawing the given triangles. Draw equilateral triangle ABC and isosceles triangle DBC, with BC as the common side.
B
/ \
/ \
/ \
A-------C
\
\
\
D
2. Since triangle ABC is equilateral, all angles in this triangle are equal to 60 degrees. Therefore, angle ABC = angle ACB = angle BAC = 60 degrees.
3. Let's find angle ABD. Since BD=BC and triangle DBC is isosceles, angle BDC = angle BCD. Given that angle BDC = 34 degrees, we can conclude that angle BCD is also 34 degrees.
4. Since the sum of angles in a triangle is 180 degrees, we can find angle CBD. Angle BCD + angle CBD + angle BDC = 180 degrees. Substituting the known values, we have 34 + angle CBD + 34 = 180. Simplifying, we get angle CBD = 112 degrees.
5. Since BC=BD and triangle CBD is isosceles, angle CBD = angle BDC. Therefore, angle BDC = 112 degrees.
6. Now, let's find angle ABD. Since angle ABD + angle BDC + angle BDA = 180 degrees, we can substitute the known values. angle ABD + 112 + 60 = 180 degrees. Simplifying, we have angle ABD = 8 degrees.
Therefore, the measure of angle ABD is 8 degrees.