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Geometry Difficulty 3.4 AMC 10/12 Find the answer Canada

In the diagram, point $D$ lies on side $BC$ of $\triangle ABC$ so that $AB=AD=CD$ .If $\angle ABC = 80\degree$ , the measure of $\angle ACD$ is

Pick one

Solutions — 2

Solution 1

Since $AB=AD$ , then $\triangle ABD$ is isosceles and $\angle ADB=\angle ABD=80\degree$ . Since $\angle BDC$ is a straight angle, then $\$\angle ADC=180\degree-\angle ADB=180\degree-80\degree=100\degree$ $.$ $\triangle$ ADC $is also isosceles (since$ AD=CD $), and so$ $\angle CAD=\angle$ ACD. The sum of the angles in

Figure for this problem $\triangle$
ADC $is$ $180\degree$ $, and so$ $\angle ADC+\angle CAD+\angle ACD=180\degree$ $or$ $100\degree+2\times\angle ACD=180\degree$ $or$ $2\times\angle ACD=80\degree$ $, and so the measure of$ $\angle$ ACD $is$ $40\degree\$.$

Solution 2

$\triangle ABD$ is isosceles with $AB=AD$ , and so $\angle ADB=\angle ABD=80\degree$ . The measure of $\angle BDC$ is 180 $\degree$ since it is a straight angle. Thus, $\$\angle ADC=180\degree-\angle ADB=180\degree-80\degree=100\degree$ $.$ $\triangle$ ADC $is isosceles with$ AD=DC $, and so$ $\angle ACD=\angle$ CAD. The sum of the three angles in

Figure for this problem $\triangle$ ADC $is$ $180\degree$ $, and so$ $\angle ACD+\angle CAD=180\degree-\angle ADC=180\degree-100\degree=80\degree$ $. Since$ $\angle ACD+\angle CAD=80\degree$ $and$ $\angle ACD=\angle$ CAD $, then$ $\angle ACD=\dfrac{80\degree}{2}=40\degree\$.$

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