Question 356·Hard·Math

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In the $xy$ -plane, point $A$ is $(0,0)$ and point $B$ is $(k,0)$ , where $k$ is a positive constant. Point $C$ is $(k,2k)$ and point $D$ is $(2k,2k)$ .

Triangles $ABC$ and $ABD$ are drawn. The measure of $\angle CAB$ is $t^\circ$ .

What is the measure of $\angle ABD$ expressed in terms of $t$ ?

Coordinates can reveal parallel segments, segments that run in the same direction, even without a diagram. Compare the right-and-up move between each pair of endpoints, then carry the given angle to a matching angle at the other point. Check the horizontal rays before setting angles equal: if one points right and the other points left, the angles add to $180^\circ$ instead.

Hints

Hint 1
From $A$ to $C$ , you move right $k$ and up $2k$ . Compare that with the move from $B$ to $D$ . Identical moves make the slanted segments parallel: they run in the same direction.
Hint 2
Parallel slanted segments make equal angles with horizontal rays that both point right. Imagine a rightward ray starting at $B$ . How does its angle with $BD$ compare with the given angle at $A$ ?
Hint 3
A linear pair is two side-by-side angles whose other sides point in opposite directions; they add to $180^\circ$ . The matching angle at $B$ uses a rightward ray, but $\angle ABD$ uses leftward $BA$ . What do their measures add to?

Step-by-step

Match the slanted directions, then use a straight angle
Step 1
Show that the slanted segments are parallel
No Desmos needed. Matching coordinate moves and a straight angle settle this. Compare the moves from each segment's starting point to its endpoint:
$\overrightarrow{AC}=(k,2k),\qquad \overrightarrow{BD}=(2k-k,2k-0)=(k,2k).$
Both move right $k$ and up $2k$ , so $AC$ and $BD$ point the same way and are parallel.
Step 2
Carry the given angle to point B
Let $E$ be a point to the right of $B$ on the $x$ -axis. The rays $AB$ and $BE$ both point right, while the parallel rays $AC$ and $BD$ point the same way. So the angle between $BD$ and $BE$ matches the given angle: $m\angle DBE=t^\circ$ .
Step 3
Identify the straight angle at B
At $B$ , $BE$ points right, but $BA$ points left. These opposite rays make $\angle DBE$ and $\angle ABD$ a linear pair: the two angles add to $180^\circ$ rather than having equal measures. So
$m\angle ABD+t^\circ=180^\circ.$
Step 4
Find the requested angle
Subtract $t^\circ$ from both sides:
$m\angle ABD=(180-t)^\circ.$
So the angle at $B$ measures $(180-t)^\circ$ . Choice C.

Question 356·Hard·Math

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