Super-Hard SAT Geometry Question 10 | Free SAT Prep

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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 $BXD$ ?

When two triangles are formed by intersecting lines, first check for vertical angles. If the question asks for similarity, use the sides that touch those vertical angles and test whether their ratios match. When two segments make up a known whole segment, write one as the total minus the other before setting up the proportion.

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Use the intersecting lines

Look at the angles at $X$ . The angle in each triangle formed by one ray from each line is a pair of vertical angles.

Match the sides around the angle

For Side-Angle-Side similarity, compare $AX$ with $XB$ and compare $XC$ with $XD$ .

Use the whole segment

If you let $XC=x$ , then use $CD=119$ to write $XD$ in terms of $x$ . Set the two required side ratios equal.

Desmos Guide

Use a ratio equation

A paper-and-pencil ratio setup is fastest here, but Desmos can verify the missing segment. Let $x$ represent $XC$ , so $XD=119-x$ .

Graph the two ratios

Enter $y=\frac{33}{44}$ and $y=\frac{x}{119-x}$ . The intersection represents the value of $x$ for which the side ratios around the vertical angles are equal.

Interpret the intersection

The intersection's $x$ -coordinate is the needed value of $XC$ . Therefore, the correct choice is A, $XC=51$ .

Step-by-step Explanation

Identify the congruent angles

Since lines $AB$ and $CD$ intersect at $X$ , angles $AXC$ and $BXD$ are vertical angles. Therefore, they are congruent.

Set the required side ratios

For Side-Angle-Side similarity using the angles at $X$ , the surrounding sides must satisfy

\frac{AX}{XB}=\frac{XC}{XD}

Because $AX=33$ and $XB=44$ , the required ratio is

\frac{XC}{XD}=\frac{33}{44}=\frac{3}{4}.

Let $XC=x$ . Since $CD=119$ , $XD=119-x$ . Thus,

\frac{x}{119-x}=\frac{3}{4}.

Cross-multiplying gives $4x=357-3x$ , so $7x=357$ .

Choose the sufficient length

Solving gives $x=51$ , so $XC=51$ . This makes $XD=68$ , and $\frac{XC}{XD}=\frac{51}{68}=\frac{3}{4}$ , matching $\frac{AX}{XB}$ . Therefore, the correct choice is A, $XC=51$ .

Strategy

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Step-by-step Explanation

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Step-by-step Explanation