Question

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

Explanation:

Step1: Recall the geometric mean theorem (altitude-on-hypotenuse theorem)

In a right - triangle \( \triangle ABC\) with the right angle at \(B\) and altitude \(BD\) to the hypotenuse \(AC\), we have three similar right - triangles: \( \triangle ADB\sim\triangle BDC\sim\triangle ABC\).

Step2: Analyze each option

• Option A:

For two triangles \( \triangle ADB\) and \( \triangle BDC\), \(\angle ADB=\angle BDC = 90^{\circ}\). \(\angle ABD+\angle CBD = 90^{\circ}\) and \(\angle C+\angle CBD=90^{\circ}\), so \(\angle ABD=\angle C\). By the AA (angle - angle) similarity criterion (\(\angle ADB=\angle BDC\) and \(\angle ABD=\angle C\)), \( \triangle ADB\sim\triangle BDC\). Congruence (\(\triangle ADB\cong\triangle BDC\)) would require side - length equality which is not guaranteed just by \(BD\) being an altitude.

• Option B:

Since \( \triangle ADB\sim\triangle BDC\) (by AA similarity as shown above).

• Option C:

From the geometric mean theorem, the correct proportion is \(\frac{AB}{BC}=\frac{BD}{DC}\) (from \( \triangle ABD\sim\triangle BCD\)), not \(\frac{AB}{AC}=\frac{BC}{BD}\).

• Option D:

\(\angle BAC\) and \(\angle BDC\): \(\angle BAC+\angle C = 90^{\circ}\) and \(\angle BDC = 90^{\circ}\). There is no reason for \(\angle BAC\cong\angle BDC\) in general.

Answer:

B. \( \triangle ADB\sim\triangle BDC\)