1998 CEMC Pascal Problems/Problem 12

Problem

In the diagram $DA = CB$ . What is the measure of $\angle DAC$ ?

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$\text{ (A) }\ 70^{\circ} \qquad\text{ (B) }\ 100^{\circ} \qquad\text{ (C) }\ 95^{\circ} \qquad\text{ (D) }\ 125^{\circ} \qquad\text{ (E) }\ 110^{\circ}$

Solution

We can use the fact that the sum of the interior angles in a triangle is $180^{\circ}$ , which gives us:

$m\angle CAB + 70 + 55 = 180$

$m\angle CAB + 125 = 180$

$m\angle CAB = 55^{\circ}$

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Since $m\angle CAB$ = $m\angle ABC$ in triangle $ABC$ , we have $AC = CB$ . However, $DA = CB$ , so $DA = AC$ from the transitive property.

This means that $m\angle ACD$ = $m\angle CDA = 40^{\circ}$ in triangle $ACD$ .

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Using the sum of the interior angles in a triangle, we have:

$m\angle DAC + 40 + 40 = 180$

$m\angle DAC + 80 = 180$

$m\angle DAC = \boxed {\textbf {(B) } 100^{\circ}}$

~anabel.disher

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1998 CEMC Pascal Problems/Problem 12

Problem

Solution