Triangle ABC is isosceles with AB=AC. D is a point on BC such that AD bisects ∠BAC. What is the measure of ∠ADB (in degrees)?
�ÚADB is a right angle
To find the measure of ∠ADB, we can use the properties of isosceles triangles and the fact that AD is an angle bisector.
Since triangle ABC is isosceles with AB = AC, we know that ∠B = ∠C. Therefore, ∠BAC is divided into two equal angles, so ∠BAC = ∠BAD + ∠DAC.
Since AD is an angle bisector, ∠BAD = ∠DAC.
Let's call ∠BAD = ∠DAC = x.
Now, we have ∠BAC = x + x = 2x.
We also know that the angles in a triangle add up to 180 degrees.
Therefore, ∠B + ∠A + ∠C = 180°. Since ∠B = ∠C, we can rewrite it as:
∠B + ∠A + ∠B = 180°
2∠B + ∠A = 180°
Since ∠B = ∠C = x, we can rewrite it as:
2x + ∠A = 180°
Now, we can substitute ∠BAC with 2x:
2x + ∠A = 180°
2x + 2x = 180°
4x = 180°
Divide both sides by 4:
x = 45°
Therefore, ∠BAD = ∠DAC = 45°.
Since AD bisects ∠BAC, ∠ADB is half of ∠BAD.
∴ ∠ADB = 45° / 2 = 22.5°
Therefore, the measure of ∠ADB is 22.5 degrees.