Question

find the lengths of segments ad and bd. then check your answers using a different method. round to the nearest tenth if necessary.
type your answers in the boxes.
ad = and bd =

Explanation:

Step1: Identify Triangle Properties

This is a right triangle \( \triangle ABC \) with \( \angle B = 90^\circ \), \( BC = 13 \), \( AC = 12 + AD \), and \( BD \perp AC \). By the geometric mean theorem (altitude-on-hypotenuse theorem), in a right triangle, the length of the altitude to the hypotenuse is the geometric mean of the lengths of the two segments it divides the hypotenuse into. Also, each leg is the geometric mean of the hypotenuse and the adjacent segment. First, find \( AB \) using Pythagoras? Wait, no, \( AC \) is the hypotenuse? Wait, no, \( \angle B \) is right, so \( AC \) is the hypotenuse? Wait, \( BC = 13 \), \( AC \) has \( AD + DC = AD + 12 \)? Wait, no, the diagram: \( A---D---C \), \( D \) between \( A \) and \( C \), \( DC = 12 \)? Wait, no, the problem says \( DC = 12 \)? Wait, the diagram shows \( DC = 12 \), \( BC = 13 \), \( \angle B = 90^\circ \), \( \angle D = 90^\circ \). So \( \triangle ABC \sim \triangle BDC \sim \triangle ADB \) (similar triangles).

By geometric mean theorem: \( BC^2 = DC \times AC \), \( AB^2 = AD \times AC \), \( BD^2 = AD \times DC \). Wait, first, find \( AC \) using \( BC^2 = DC \times AC \). So \( 13^2 = 12 \times AC \)? Wait, no, \( BC = 13 \), \( DC = 12 \), so \( 13^2 = 12 \times AC \)? Wait, \( 169 = 12 \times AC \), so \( AC = \frac{169}{12} \approx 14.08 \)? No, that can't be. Wait, maybe I mixed up. Wait, \( \triangle ABC \) is right-angled at \( B \), \( BD \perp AC \). So the correct geometric mean: \( BD^2 = AD \times DC \), \( AB^2 = AD \times AC \), \( BC^2 = DC \times AC \). So \( BC^2 = DC \times AC \implies 13^2 = 12 \times AC \implies AC = \frac{169}{12} \approx 14.08 \)? No, that's wrong. Wait, maybe \( AC \) is the hypotenuse, \( BC = 13 \), \( AB \) is another leg, \( AC = 12 + AD \). Wait, no, the diagram: \( A---D---C \), \( DC = 12 \), \( BC = 13 \), \( \angle B = 90^\circ \), \( \angle D = 90^\circ \). So \( \triangle BDC \) is right-angled at \( D \), \( \triangle ABC \) right-angled at \( B \). So \( \triangle ABC \sim \triangle BDC \) (AA similarity: \( \angle C \) common, \( \angle B = \angle D = 90^\circ \)). So \( \frac{BC}{AC} = \frac{DC}{BC} \implies BC^2 = DC \times AC \implies 13^2 = 12 \times AC \implies AC = \frac{169}{12} \approx 14.08 \). Then \( AD = AC - DC = \frac{169}{12} - 12 = \frac{169 - 144}{12} = \frac{25}{12} \approx 2.1 \)? No, that doesn't make sense. Wait, maybe I got the sides wrong. Wait, maybe \( AC = 12 \), \( BC = 13 \)? No, the diagram shows \( DC = 12 \), \( BC = 13 \). Wait, maybe the triangle is \( \triangle ABC \) with \( AC = 12 \), \( BC = 13 \), right-angled at \( B \)? No, that would make \( AB = \sqrt{13^2 - 12^2} = 5 \). Ah! Wait, that's the key. If \( AC \) is the hypotenuse? No, \( \angle B = 90^\circ \), so \( AC \) is the hypotenuse. Wait, \( AB^2 + BC^2 = AC^2 \). If \( BC = 13 \), \( AB = 5 \), \( AC = 12 \)? No, \( 5^2 + 12^2 = 13^2 \). Oh! Wait, I had the legs reversed. \( \triangle ABC \) is right-angled at \( B \), \( AB = 5 \), \( BC = 12 \), \( AC = 13 \)? No, \( 5-12-13 \) triangle. Wait, the diagram: \( A---D---C \), \( D \) on \( AC \), \( BD \perp AC \), \( DC = 12 \)? No, that's not. Wait, the problem's diagram: \( AC \) is the hypotenuse, length \( AC \), \( BC = 13 \), \( AB \) is a leg, \( BD \) is altitude. Wait, the 5-12-13 triangle: right-angled at \( B \), \( AB = 5 \), \( BC = 12 \), \( AC = 13 \). But in the diagram, \( DC = 12 \), \( BC = 13 \). Wait, maybe the diagram has \( DC = 12 \), \( BC = 13 \), so \( \triangle BDC \) has \( DC = 12 \), \( BC = 13 \), right-angled at \( D \), so \( BD = \sqrt{13^2 - 12^2} = 5 \). Then, by similar triangles, \( \triangle ADB \sim \triangle BDC \), so \( \frac{AD}{BD} = \frac{BD}{DC} \implies BD^2 = AD \times DC \implies 5^2 = AD \times 12 \implies AD = \frac{25}{12} \approx 2.1 \)? No, that's not. Wait, no, the 5-12-13 triangle: right-angled at \( B \), \( AB = 5 \), \( BC = 12 \), \( AC = 13 \). Then \( BD \) is the altitude to \( AC \). So \( BD = \frac{AB \times BC}{AC} = \frac{5 \times 12}{13} = \frac{60}{13} \approx 4.6 \). And \( AD = \frac{AB^2}{AC} = \frac{25}{13} \approx 1.9 \)? Wait, no, \( AD = \frac{AB^2}{AC} = \frac{25}{13} \approx 1.9 \), \( DC = \frac{BC^2}{AC} = \frac{144}{13} \approx 11.1 \). But the problem says \( DC = 12 \). So maybe the diagram has \( DC = 12 \), \( BC = 13 \), so \( AC = AD + 12 \), \( BD \perp AC \). Let's use Pythagoras in \( \triangle BDC \): \( BD^2 + 12^2 = 13^2 \implies BD^2 = 169 - 144 = 25 \implies BD = 5 \). Then, in \( \triangle ABD \), \( AD^2 + BD^2 = AB^2 \), and in \( \triangle ABC \), \( AB^2 + 13^2 = (AD + 12)^2 \). Substitute \( AB^2 = AD^2 + 25 \) into the second equation: \( AD^2 + 25 + 169 = AD^2 + 24AD + 144 \implies 194 = 24AD + 144 \implies 24AD = 50 \implies AD = \frac{50}{24} = \frac{25}{12} \approx 2.1 \). Wait, but \( BD = 5 \). Let's check with geometric mean: \( BD^2 = AD \times DC \implies 25 = \frac{25}{12} \times 12 \implies 25 = 25 \), which works. So \( AD = \frac{25}{12} \approx 2.1 \)? No, \( \frac{25}{12} \approx 2.1 \), but \( BD = 5 \). Wait, the problem's diagram: maybe \( AC = 12 \), \( BC = 13 \), but that's not. Wait, the user's diagram: \( A---D---C \), \( D \) between \( A \) and \( C \), \( DC = 12 \), \( BC = 13 \), \( \angle B = 90^\circ \), \( \angle D = 90^\circ \). So \( \triangle BDC \): \( BD \perp AC \), \( DC = 12 \), \( BC = 13 \). So \( BD = \sqrt{13^2 - 12^2} = 5 \). Then, by geometric mean, \( BD^2 = AD \times DC \implies 25 = AD \times 12 \implies AD = \frac{25}{12} \approx 2.1 \). Wait, but \( AC = AD + DC = \frac{25}{12} + 12 = \frac{25 + 144}{12} = \frac{169}{12} \approx 14.08 \). Then in \( \triangle ABC \), \( AB^2 + BC^2 = AC^2 \implies AB^2 + 169 = (\frac{169}{12})^2 \implies AB^2 = \frac{28561}{144} - 169 = \frac{28561 - 24336}{144} = \frac{4225}{144} \implies AB = \frac{65}{12} \approx 5.4 \). Then check \( \triangle ABD \): \( AD^2 + BD^2 = (\frac{25}{12})^2 + 25 = \frac{625}{144} + \frac{3600}{144} = \frac{4225}{144} = AB^2 \), which works. So \( AD = \frac{25}{12} \approx 2.1 \), \( BD = 5 \). Wait, but the problem says "round to the nearest tenth if necessary". \( \frac{25}{12} \approx 2.1 \), \( BD = 5 \).

Step2: Verify with Another Method

Use area of \( \triangle ABC \). Area via \( \frac{1}{2} \times AB \times BC = \frac{1}{2} \times \frac{65}{12} \times 13 = \frac{845}{24} \). Area via \( \frac{1}{2} \times AC \times BD = \frac{1}{2} \times \frac{169}{12} \times 5 = \frac{845}{24} \), which matches. Also, \( AD = \frac{AB^2}{AC} = \frac{(\frac{65}{12})^2}{\frac{169}{12}} = \frac{4225}{144} \times \frac{12}{169} = \frac{4225}{12 \times 169} = \frac{25}{12} \approx 2.1 \), which matches.

Answer:

\( AD = \frac{25}{12} \approx 2.1 \) and \( BD = 5 \) (or \( AD \approx 2.1 \), \( BD = 5 \))