In right triangle , the measure of angle is and has length 15. Point lies on segment with . Which choice lists all statements that must be true? I. II. III.

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Super-Hard SAT Math Question 4 | Free SAT Prep | Aniko

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Question 4·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In right triangle $\triangle ABC$ , the measure of angle $B$ is $90^\circ$ and $\overline{AB}$ has length 15. Point $D$ lies on segment $\overline{AC}$ with $BD\perp AC$ .

Which choice lists all statements that must be true?

I. $BD^2 = AD\cdot DC$

II. $\dfrac{AD}{DC}=\left(\dfrac{AB}{BC}\right)^2$

III. $\angle ABD = \angle BDC$

When a right triangle includes an altitude from the right angle to the hypotenuse, immediately look for similar triangles ( $\triangle ABD$ , $\triangle CBD$ , and $\triangle ABC$ ). Use similarity to derive the standard consequences: the altitude is the geometric mean ( $BD^2=AD\cdot DC$ ) and each leg is the geometric mean of the hypotenuse and its adjacent projection (which lets you form ratios like $AD/DC=(AB/BC)^2$ ). For any angle-equality claim, compute what one angle must be from perpendicularity (often $90^\circ$ ) and check whether the other angle is actually forced to match.

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Recognize what $BD\perp AC$ implies

If $\angle B=90^\circ$ and $BD$ is perpendicular to $AC$ , then $BD$ is the altitude from the right angle to the hypotenuse. What similarity relationships does that create?

Target statement I with similar triangles

Try proving $BD^2=AD\cdot DC$ by comparing $\triangle ABD$ and $\triangle CBD$ . Can you set up a proportion that has $BD$ in both numerator and denominator?

Relate $AD$ and $DC$ to $AB$ and $BC$

Use similarity between each small triangle and the original triangle to get equations like $AB^2=AD\cdot AC$ and $BC^2=DC\cdot AC$ . Then divide to eliminate $AC$ .

Desmos Guide

Set up a right triangle with a slider

Create a slider $c>0$ . Define $B=(0,0)$ , $A=(15,0)$ , and $C=(0,c)$ . Then $\angle B$ is $90^\circ$ and $AB=15$ .

Create point $D$ as the foot of the perpendicular from $B$ to $AC$

In Desmos, create point $D$ as the projection of $B$ onto line $AC$ (you can use the built-in construction for the foot of a perpendicular, or compute the projection with vectors).

This guarantees $D$ lies on segment $AC$ and $BD\perp AC$ .

Verify statement I numerically

Compute $AD$ , $DC$ , and $BD$ , then compare $BD^2$ to $AD\cdot DC$ for several values of $c$ . They should match (up to minor rounding).

Verify statement II numerically

Compute $AD/DC$ and compare it to $(AB/BC)^2$ (here $AB=15$ and $BC=c$ ). They should match for several values of $c$ .

Test statement III with a sample

Measure $\angle BDC$ (it should be $90^\circ$ ) and measure $\angle ABD$ (it will generally not be $90^\circ$ ). Since these angles are not always equal, statement III is not a must.

Step-by-step Explanation

Use the altitude-to-hypotenuse similarity fact

Since $\angle B = 90^\circ$ and $BD\perp AC$ , point $D$ is the foot of the altitude from the right angle to the hypotenuse.

This creates two smaller right triangles, $\triangle ABD$ and $\triangle CBD$ , each with a right angle at $D$ . By AA similarity (matching one acute angle as well),

\triangle ABD \sim \triangle ABC \quad\text{and}\quad \triangle CBD \sim \triangle ABC,

so in particular $\triangle ABD\sim\triangle CBD$ as well.

Prove statement I: $BD^2 = AD\cdot DC$

From $\triangle ABD\sim\triangle CBD$ , corresponding sides give

\frac{BD}{AD}=\frac{DC}{BD}.

Multiplying both sides by $AD\cdot BD$ yields

BD^2=AD\cdot DC,

so statement I must be true.

Prove statement II using projections on the hypotenuse

From $\triangle ABD\sim\triangle ABC$ ,

\frac{AB}{AC}=\frac{AD}{AB}\quad\Rightarrow\quad AB^2=AD\cdot AC.

From $\triangle CBD\sim\triangle ABC$ ,

\frac{BC}{AC}=\frac{DC}{BC}\quad\Rightarrow\quad BC^2=DC\cdot AC.

Divide the two equations:

\frac{AB^2}{BC^2}=\frac{AD\cdot AC}{DC\cdot AC}=\frac{AD}{DC}

\frac{AD}{DC}=\left(\frac{AB}{BC}\right)^2,

and statement II must be true.

Check statement III

Because $BD\perp AC$ and $DC$ lies on line $AC$ , angle $\angle BDC$ is a right angle:

\angle BDC = 90^\circ.

But $\angle ABD$ is the angle between $\overline{BA}$ and $\overline{BD}$ , and there is no reason $BD$ must be perpendicular to $AB$ . So $\angle ABD$ is not forced to equal $90^\circ$ , meaning statement III does not have to be true.

Thus, the correct choice is I and II only.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation