Question 10·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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Lines $AB$ and $CD$ intersect at point $X$ . Point $X$ lies between $A$ and $B$ and between $C$ and $D$ . The lengths satisfy $AX=33$ , $XB=44$ , and $CD=119$ .

Which choice, if true, is sufficient to prove that triangle $AXC$ is similar to triangle $BXD$ ?

Intersecting lines give vertical angles, equal opposite angles at the crossing. To prove similarity from that one equal angle, use SAS similarity: the two pairs of sides touching it need the same ratio. If a full segment is given, split it into the two pieces; the full segment isn't a side of either triangle. Match sides before you compare their ratios.

Hints

Hint 1
At intersecting lines, vertical angles sit opposite each other and have equal measures. Which angles at $X$ belong to the two triangles? Their equality supplies the angle needed for SAS similarity.
Hint 2
For corresponding sides, pair vertices in the order of the triangle names: $A$ with $B$ , $X$ with $X$ , and $C$ with $D$ . Which two side pairs touch the equal angles?
Hint 3
The length $CD$ covers both pieces separated by $X$ . If $x$ means $XC$ , write $XD$ in terms of $x$ . Then the SAS similarity test compares $AX/XB$ with $XC/XD$ .

Step-by-step

Use the equal angle and two side ratios
Step 1
Find the equal angles
At the crossing, $\angle AXC$ and $\angle BXD$ sit opposite each other. Vertical angles are opposite angles formed by crossing lines, so these angles are equal.
Step 2
Match the sides around those angles
The triangle names pair $A$ with $B$ , $X$ with $X$ , and $C$ with $D$ . This correspondence, or matching, pairs the sides touching the equal angles: $AX$ with $BX$ , and $XC$ with $XD$ .
Step 3
Write the other piece of $CD$
Let $x=XC$ . Because $X$ lies between $C$ and $D$ , the two pieces add to the whole:
$119=x+XD.$
Subtract $x$ to express the other piece:
$XD=119-x.$
Step 4
Build the SAS similarity equation
For SAS similarity, the equal angle must lie between two pairs of sides with matching ratios. Keep the first triangle over the second in both ratios:
$\frac{AX}{BX}=\frac{XC}{XD}.$
Substitute the lengths and $XD$ from the previous step:
$\frac{33}{44}=\frac{x}{119-x}.$
Don't use $CD$ as $XD$ : $CD$ is the whole segment, not a side of triangle $BXD$ .
Step 5
Find the length that makes the ratios match
Type $\frac{33}{44}\sim\frac{x_{1}}{119-x_{1}}\{0<x_{1}<119\}$ in Desmos. Use $x_1$ for the unknown length $XC$ ; $\sim$ tells Desmos to find the value that makes the sides agree, and the braces keep it between $0$ and $119$ . Under PARAMETERS, Desmos shows $x_1=51$ . So $XC=51$ makes the sides proportional, and the equal vertical angles prove the triangles similar by SAS. Choice A.

Question 10·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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