Question

practice 9 (from unit 2, lesson 6)
triangles acd and bcd are isosceles. angle dbc has a measure of 110° and angle bda has a measure of 22°.
$\overline{ad} \cong \overline{ac}$
$\overline{bd} \cong \overline{bc}$
find the measure of angle bac.
type your answer in the box.
angle bac has a measure of degrees.

Explanation:

Step1: Find angle BDC in triangle DBC

In isosceles triangle \(DBC\) (since \(BD\cong BC\)), the base angles are equal. The sum of angles in a triangle is \(180^\circ\). Given \(\angle DBC = 110^\circ\), let \(\angle BDC=\angle BCD = x\). Then \(x + x+ 110^\circ=180^\circ\), so \(2x = 70^\circ\), and \(x = 35^\circ\). So \(\angle BDC = 35^\circ\).

Step2: Find angle ADC

We know \(\angle BDA = 22^\circ\) and \(\angle BDC = 35^\circ\), so \(\angle ADC=\angle BDA+\angle BDC = 22^\circ + 35^\circ=57^\circ\).

Step3: Find angle ACD in isosceles triangle ACD

In isosceles triangle \(ACD\) (since \(AD\cong AC\)), \(\angle ADC=\angle ACD = 57^\circ\) (base angles of isosceles triangle are equal).

Step4: Find angle DAC in triangle ACD

The sum of angles in triangle \(ACD\) is \(180^\circ\). Let \(\angle DAC = y\). Then \(y+57^\circ + 57^\circ=180^\circ\), so \(y = 180^\circ- 114^\circ = 66^\circ\). Wait, no, wait. Wait, actually, we need to find \(\angle BAC\). Wait, maybe a better way: Let's find \(\angle ACB\) first. Wait, \(\angle ACB=\angle ACD-\angle BCD\). We know \(\angle ACD = 57^\circ\) and \(\angle BCD = 35^\circ\), so \(\angle ACB = 57^\circ - 35^\circ=22^\circ\). Wait, no, maybe I made a mistake. Wait, let's re - evaluate.

Wait, in triangle \(DBC\), \(BD = BC\), so \(\angle BDC=\angle BCD=(180 - 110)/2 = 35^\circ\). Then \(\angle ADC=\angle ADB+\angle BDC = 22 + 35=57^\circ\). In triangle \(ADC\), \(AD = AC\), so \(\angle ADC=\angle ACD = 57^\circ\). Then \(\angle DAC=180 - 57 - 57 = 66^\circ\). Now, let's look at triangle \(ABC\) and \(ABD\). Wait, maybe we can use the fact that \(AD = AC\) and \(BD = BC\), and \(AB\) is common? Wait, no. Wait, let's find \(\angle DAB\) first. In triangle \(ABD\), we know \(\angle ADB = 22^\circ\), and we can find \(\angle ABD\). \(\angle ABD = 180^\circ-\angle DBC=180 - 110 = 70^\circ\) (since \(\angle DBC\) and \(\angle ABD\) are supplementary). Then in triangle \(ABD\), \(\angle DAB=180 - 22 - 70 = 88^\circ\). Then \(\angle BAC=\angle DAC-\angle DAB\)? No, that's not right. Wait, no, \(\angle DAC\) is the angle at \(A\) in triangle \(ADC\), and \(\angle DAB\) is the angle at \(A\) in triangle \(ABD\). Wait, maybe the correct approach is:

Since \(AD = AC\) and \(BD = BC\), and \(AB\) is common. Wait, let's find \(\angle ABC\). \(\angle ABC = 180^\circ-\angle DBC = 70^\circ\) (linear pair). In triangle \(ABC\), \(BC = BD\), \(AC = AD\), and \(AB\) is common? Wait, no. Wait, let's find \(\angle BAC\) by considering triangle \(ABC\) and triangle \(ABD\).

Wait, another way: We know \(\angle ADC = 57^\circ\), \(AD = AC\), so \(\angle DAC=180 - 2\times57 = 66^\circ\). Now, let's find \(\angle DAB\). In triangle \(ABD\), angles are \(\angle ADB = 22^\circ\), \(\angle ABD = 180 - 110 = 70^\circ\), so \(\angle DAB=180-(22 + 70)=88^\circ\). Wait, that can't be, because \(88>66\). So I must have messed up.

Wait, no, \(\angle ABD\) is not \(180 - 110\). Wait, \(\angle DBC = 110^\circ\), so \(\angle ABD = 180^\circ - 110^\circ=70^\circ\) is correct (since they are adjacent angles on a straight line? Wait, no, the points are \(D - B - A\)? Wait, looking at the diagram, \(B\) is inside the triangle. So \(\angle ABD\) and \(\angle DBC\) are not supplementary. Oh! That's my mistake. So \(\angle DBC = 110^\circ\), so in triangle \(DBC\), \(BD = BC\), so \(\angle BDC=\angle BCD = 35^\circ\) as before. Then \(\angle ADB = 22^\circ\), so \(\angle ADC=\angle ADB+\angle BDC = 22 + 35 = 57^\circ\). In triangle \(ADC\), \(AD = AC\), so \(\angle ACD=\angle ADC = 57^\circ\). Now, let's find \(\angle ACB\). \(\angle ACB=\angle ACD-\angle BCD=57 - 35 = 22^\circ\). Now, in triangle \(ABC\) and triangle \(ABD\), \(AD = AC\), \(BD = BC\), \(AB = AB\), so triangle \(ABD\cong triangle ABC\) (SSS). Wait, \(AD = AC\), \(BD = BC\), \(AB\) is common. So \(\angle BAD=\angle BAC\)? No, wait, if \(AD = AC\), \(BD = BC\), \(AB = AB\), then \(\triangle ABD\cong\triangle ABC\) (SSS). So \(\angle ADB=\angle ACB = 22^\circ\), which matches. Now, in triangle \(ADC\), \(AD = AC\), so it's isosceles with \(\angle ADC=\angle ACD = 57^\circ\), so \(\angle DAC=180 - 57\times2=66^\circ\). Since \(\triangle ABD\cong\triangle ABC\), \(\angle BAD=\angle BAC\), and \(\angle DAC=\angle BAD+\angle BAC = 2\angle BAC\). So \(\angle BAC=\angle DAC/2 = 66/2 = 33^\circ\). Wait, no, wait. Wait, \(\angle DAC=\angle DAB+\angle BAC\), and if \(\triangle ABD\cong\triangle ABC\), then \(\angle DAB=\angle BAC\), so \(\angle DAC = 2\angle BAC\). We found \(\angle DAC = 180 - 57 - 57=66^\circ\), so \(\angle BAC = 33^\circ\). Wait, let's check again.

Wait, let's start over.

1. In \(\triangle DBC\), \(BD = BC\), so it's isosceles. \(\angle DBC = 110^\circ\), so \(\angle BDC=\angle BCD=(180 - 110)/2 = 35^\circ\).

2. \(\angle ADB = 22^\circ\), so \(\angle ADC=\angle ADB+\angle BDC=22 + 35 = 57^\circ\).

3. In \(\triangle ADC\), \(AD = AC\), so it's isosceles with \(\angle ACD=\angle ADC = 57^\circ\).

4. The sum of angles in \(\triangle ADC\): \(\angle DAC=180-(57 + 57)=66^\circ\).

5. Now, consider \(\triangle ABD\) and \(\triangle ABC\): \(AD = AC\), \(BD = BC\), \(AB = AB\), so \(\triangle ABD\cong\triangle ABC\) (SSS). Therefore, \(\angle BAD=\angle BAC\).

6. Since \(\angle DAC=\angle BAD+\angle BAC = 2\angle BAC\), then \(\angle BAC=\frac{\angle DAC}{2}=\frac{66^\circ}{2}=33^\circ\).

Answer:

33