Question

1. given that $\triangle abc$ is an isosceles triangle and $\overrightarrow{ad}$ and $\overrightarrow{cd}$ are angle bisectors, what is $m\angle adc$?

Explanation:

Step1: Analyze the isosceles triangle

In \(\triangle ABC\), since it's isosceles with vertex angle (let's assume \(\angle BAC\) or the top angle) given (from the diagram, the top angle seems to be \(40^\circ\), so \(\angle B = \angle C\) in a typical isosceles triangle? Wait, no, in an isosceles triangle, the base angles are equal. Wait, actually, if the vertex angle at \(B\) (the top) is \(40^\circ\), then the base angles \(\angle BAC\) and \(\angle BCA\) would be equal. Wait, let's correct: The sum of angles in a triangle is \(180^\circ\). So if \(\angle ABC = 40^\circ\), then \(\angle BAC + \angle BCA = 180 - 40 = 140^\circ\), and since \(\triangle ABC\) is isosceles with \(AB = BC\)? No, wait, the triangle is labeled with \(A\) and \(C\) at the base, \(B\) at the top. So \(AB = BC\)? No, \(AB = AC\) would make the base \(BC\). Wait, the diagram shows \(A\) and \(C\) at the bottom, \(B\) at the top, and \(D\) inside. So \(\triangle ABC\) is isosceles with \(AB = AC\), so \(\angle ABC = \angle ACB\)? Wait, no, if \(AB = AC\), then the base is \(BC\), and the base angles are \(\angle ABC\) and \(\angle ACB\), so they are equal. Wait, the top angle is \(\angle BAC = 40^\circ\), so then \(\angle ABC + \angle ACB = 180 - 40 = 140^\circ\), so each base angle is \(70^\circ\) (since \(140/2 = 70\)).

Step2: Use angle bisectors

\(AD\) and \(CD\) are angle bisectors, so they split \(\angle BAC\) and \(\angle BCA\) into two equal parts. Wait, no: \(\angle BAC\) is at \(A\), so \(AD\) bisects \(\angle BAC\), and \(CD\) bisects \(\angle BCA\). Wait, no, the angle bisectors are \(AD\) (bisecting \(\angle BAC\)) and \(CD\) (bisecting \(\angle BCA\))? Wait, no, looking at the diagram, \(AD\) is from \(A\) to \(D\), \(CD\) is from \(C\) to \(D\), so \(AD\) bisects \(\angle BAC\) and \(CD\) bisects \(\angle BCA\)? Wait, no, actually, in the diagram, \(D\) is inside the triangle, so \(AD\) bisects \(\angle BAD\) (wait, no, \(\angle BAC\)) and \(CD\) bisects \(\angle BCA\). Wait, let's re-express:

If \(\angle BAC = 40^\circ\), then \(AD\) bisects it, so \(\angle DAC = \frac{40^\circ}{2} = 20^\circ\)? Wait, no, that can't be. Wait, maybe the vertex angle is at \(B\), so \(\angle ABC = 40^\circ\), then the base angles \(\angle BAC\) and \(\angle BCA\) are equal, so \(\angle BAC = \angle BCA = \frac{180 - 40}{2} = 70^\circ\) each. Then \(AD\) bisects \(\angle BAC\) (so \(\angle BAD = \angle DAC = 35^\circ\)) and \(CD\) bisects \(\angle BCA\) (so \(\angle BCD = \angle DCA = 35^\circ\)). Then in \(\triangle ADC\), the sum of angles is \(180^\circ\), so \(\angle ADC = 180 - \angle DAC - \angle DCA\). Wait, \(\angle DAC = 35^\circ\), \(\angle DCA = 35^\circ\), so \(\angle ADC = 180 - 35 - 35 = 110^\circ\)? Wait, no, wait: Wait, if \(\angle BAC = 70^\circ\), then bisecting it gives \(35^\circ\) for \(\angle DAC\), and \(\angle BCA = 70^\circ\), bisecting it gives \(35^\circ\) for \(\angle DCA\). Then in \(\triangle ADC\), angles sum to \(180\), so \(180 - 35 - 35 = 110\). Wait, but another way: The formula for the angle formed by two angle bisectors in a triangle: The measure of the angle between the bisectors is \(90^\circ + \frac{1}{2}\) the measure of the original vertex angle. Wait, if the original vertex angle (at \(B\)) is \(40^\circ\), then the angle at \(D\) ( \(\angle ADC\)) would be \(90 + \frac{40}{2} = 90 + 20 = 110^\circ\). Yes, that matches. So the formula is: In a triangle, if two angle bisectors are drawn from the base angles, the angle between them is \(90^\circ + \frac{1}{2}\) the vertex angle. So here, vertex angle at \(B\) is \(40^\circ\), so \(\angle ADC = 90 + \frac{40}{2} = 110^\circ\).

Step3: Verify with angle sum

Sum of angles in \(\triangle ABC\): \(\angle B = 40^\circ\), \(\angle A = \angle C = 70^\circ\) (since \(180 - 40 = 140\), divided by 2). Then \(AD\) bisects \(\angle A\) (70°) into 35°, \(CD\) bisects \(\angle C\) (70°) into 35°. Then in \(\triangle ADC\), angles are \(\angle DAC = 35^\circ\), \(\angle DCA = 35^\circ\), so \(\angle ADC = 180 - 35 - 35 = 110^\circ\). Correct.

Answer:

\(110^\circ\)