Question

which statement is necessarily true if \\(\overline{bd}\\) is an altitude to the hypotenuse of right \\(\triangle abc\\)?
\\(\bigcirc\\) a. \\(\triangle adb \cong \triangle bdc\\)
\\(\bigcirc\\) b. \\(\triangle adb \sim \triangle bdc\\)
\\(\bigcirc\\) c. \\(\frac{ab}{bc} = \frac{ac}{bd}\\)
\\(\bigcirc\\) d. \\(\angle bac \cong \angle bdc\\)

Explanation:

Step1: Recall properties of right triangles with altitude to hypotenuse

In a right triangle, when an altitude is drawn to the hypotenuse, the two smaller triangles formed are similar to the original triangle and to each other. So, for right $\triangle ABC$ with right angle at $B$ and altitude $BD$ to hypotenuse $AC$, we have $\triangle ADB \sim \triangle ABC$ and $\triangle BDC \sim \triangle ABC$, which implies $\triangle ADB \sim \triangle BDC$.

Step2: Analyze each option

• Option A: $\triangle ADB \cong \triangle BDC$ (congruent) is not necessarily true. Congruence requires all corresponding sides and angles to be equal, but here the triangles are similar (proportional sides, equal angles) but not necessarily congruent unless the original triangle is isoceles right triangle, which is not given.
• Option B: $\triangle ADB \sim \triangle BDC$ (similar) is true by the geometric mean theorem (or altitude-on-hypotenuse theorem) which states that the two smaller triangles formed by the altitude to the hypotenuse of a right triangle are similar to each other and to the original triangle.
• Option C: $\frac{AB}{BC} = \frac{AC}{BD}$ - Let's check the ratios. From similar triangles, we have $\frac{AB}{AC}=\frac{AD}{AB}$ and $\frac{BC}{AC}=\frac{CD}{BC}$, and $\frac{AB}{BC}=\frac{AD}{BD}=\frac{BD}{CD}$. The ratio $\frac{AB}{BC} = \frac{AC}{BD}$ would imply $AB \cdot BD = BC \cdot AC$, which is not a standard ratio from the similar triangles. So this is not necessarily true.
• Option D: $\angle BAC \cong \angle BDC$ - $\angle BAC$ is an acute angle in $\triangle ABC$, $\angle BDC$ is a right angle (since $BD$ is altitude, $\angle BDC = 90^\circ$), and $\angle BAC$ is not necessarily a right angle, so they can't be congruent.

Answer:

B. $\triangle ADB \sim \triangle BDC$