Super-Hard SAT Geometry Question 13 | Free SAT Prep

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Question 13·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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In triangle $ABC$ , $AB=AC$ . Point $D$ lies on segment $BC$ , and points $A$ , $D$ , and $E$ are collinear, with $D$ between $A$ and $E$ . Also, $AD=DC$ .

The measure of $\angle CAD$ is $(4x+4)^\circ$ , and the measure of $\angle CDE$ is $(6x+26)^\circ$ .

What is the measure, in degrees, of $\angle BAD$ ?

Your answer

For a problem with an isosceles triangle inside another isosceles triangle, first use the smaller triangle and any exterior angle to find a numerical angle. Then transfer that angle to the larger triangle using collinearity and equal base angles. Finally, use the triangle angle sum and subtract the known portion of the vertex angle.

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Identify an isosceles triangle

Use $AD=DC$ to compare $\angle CAD$ and $\angle ACD$ .

Use the extended side

Since $A$ , $D$ , and $E$ are collinear, $\angle CDE$ is an exterior angle of triangle $ADC$ . Write it as the sum of the two remote interior angles.

Connect the two triangles

After finding $\angle ACD$ , use the fact that $D$ lies on $BC$ and that $AB=AC$ before finding the angle at $A$ .

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Verify the equation for $x$

Algebra is faster for this problem. To verify the value of $x$ , graph $y=6x+26$ and $y=2(4x+4)$ . The $x$ -coordinate of their intersection is the value that makes the exterior-angle relationship true.

Check the final angle calculation

After obtaining the value of $x$ , enter $a=4x+4$ using that value. Then enter $180-3a$ . This represents the full angle at $A$ of triangle $ABC$ , minus $\angle CAD$ .

Step-by-step Explanation

Use the exterior angle

Since $AD=DC$ , triangle $ADC$ is isosceles, so

\angle CAD=\angle ACD.

Because $A$ , $D$ , and $E$ are collinear, $\angle CDE$ is an exterior angle of triangle $ADC$ . It equals the sum of the two remote interior angles:

6x+26=2(4x+4).

Solving gives $x=9$ .

Find the base angles of triangle $ABC$

Substitute $x=9$ into $\angle CAD$ :

\angle CAD=4(9)+4=40.

Thus $\angle ACD=40$ . Since $D$ lies on segment $BC$ , rays $CD$ and $CB$ are the same ray, so $\angle ACB=40$ .

Because $AB=AC$ , triangle $ABC$ is isosceles, and $\angle ABC=\angle ACB=40$ .

Subtract to find the requested angle

The angle sum of triangle $ABC$ gives

\angle BAC=180-40-40=100.

Segment $AD$ divides $\angle BAC$ into $\angle BAD$ and $\angle CAD$ . Therefore,

\angle BAD=100-40=60.

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Step-by-step Explanation

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