Question

in the diagram, dc is 10 units and bc is 6 units. what is the length of segment ac? 6 units 8 units 12 units 16 units

Explanation:

Step1: Identify triangle type

Triangle \(DBC\) is right - angled at \(B\) (since there is a right - angle symbol at \(B\)). We can use the Pythagorean theorem to find the length of \(DB\). The Pythagorean theorem states that for a right - triangle with hypotenuse \(c\) and legs \(a\) and \(b\), \(c^{2}=a^{2}+b^{2}\). In right - triangle \(DBC\), \(DC = 10\) (hypotenuse) and \(BC = 6\) (one leg). Let \(DB=x\). Then \(x^{2}+6^{2}=10^{2}\).

Step2: Solve for \(DB\)

\[ \begin{align*} x^{2}+36&=100\\ x^{2}&=100 - 36\\ x^{2}&=64\\ x&=\sqrt{64}\\ x& = 8 \end{align*} \]
From the diagram, we can see that \(B\) is the mid - point of \(AC\) (since the segments \(AB\) and \(BC\) are marked as equal, and also \(B\) is the mid - point of \(DE\) but for \(AC\), the markings show \(AB = BC\) in terms of the perpendicular bisector or congruent segments). Also, since \(B\) is the mid - point of \(AC\) and \(DB = 8\), and from the congruency or the properties of the perpendicular bisector (the line \(l\) is a perpendicular bisector of \(AC\) as it is perpendicular at \(B\) and \(AB = BC\)), triangle \(ABD\) and triangle \(CBD\) are congruent (by SAS congruence: \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), \(DB = DB\)). So \(AC = 2\times DC\)? No, wait. Wait, since \(B\) is the mid - point of \(AC\) and we found \(DB = 8\), but actually, looking at the right - triangle \(DBC\), we found \(DB = 8\), and since \(AB=DB\) (because of the congruency or the fact that the line \(l\) is a perpendicular bisector and the triangles are congruent), and \(AC = 2\times AB\), and \(AB = DB = 8\)? Wait, no. Wait, the key is that \(B\) is the mid - point of \(AC\) (from the markings on the diagram, the two segments of \(AC\) are equal) and triangle \(DBC\) is right - angled. We found \(DB = 8\), and since \(AB = DC\)? No, wait, let's re - examine. The line \(l\) is a perpendicular bisector of \(AC\), so \(AB = BC\) and \(\angle ABC = 90^{\circ}\)? No, the right angle is at \(B\) between \(DE\) and \(AC\). So \(DB\) is perpendicular to \(AC\) at \(B\), so \(AB = BC\) (since \(B\) is the mid - point as per the markings on \(AC\)). In right - triangle \(DBC\), \(DC = 10\), \(BC = 6\), so \(DB=\sqrt{10^{2}-6^{2}}=\sqrt{100 - 36}=\sqrt{64}=8\). Now, since \(B\) is the mid - point of \(AC\) and \(AB = DB\) (because of the congruency of triangles \(ABD\) and \(CBD\) as \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), \(DB = DB\)), so \(AB = 8\), then \(AC=AB + BC=8 + 8 = 16\)? Wait, no, that's wrong. Wait, no, \(BC = 6\)? Wait, no, the problem says \(BC = 6\) units. Wait, I made a mistake. Wait, the right - triangle is \(DBC\), with \(DC = 10\), \(BC = 6\), so \(DB=\sqrt{10^{2}-6^{2}} = 8\). Now, since \(B\) is the mid - point of \(AC\) (from the markings on \(AC\), the two parts are equal), and also, since \(DB\) is perpendicular to \(AC\), and if we consider that \(AC\) is bisected by \(DE\) at \(B\), and \(AB = DC\)? No, no. Wait, the correct approach: Since \(B\) is the mid - point of \(AC\), \(AC = 2\times AB\). And in right - triangle \(DBC\), \(DB = 8\), and since \(AB = DB\) (because the triangles \(ABD\) and \(CBD\) are congruent, as \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), \(DB = DB\)), so \(AB = 8\), then \(AC=2\times AB = 16\)? Wait, no, \(BC = 6\), but if \(AB = 8\), then \(BC\) should be \(8\) if \(B\) is the mid - point. Wait, I misread the diagram. The markings on \(AC\) show that \(AB = BC\), so \(B\) is the mid - point, so \(AB = BC\). The right - triangle is \(DBC\), with \(DC = 10\), \(BC = 6\)? No, that can't be. Wait, no, maybe \(BC\) is \(6\), and \(DC\) is \(10\), and the right angle is at \(B\), so \(DB=\sqrt{10^{2}-6^{2}} = 8\). And since \(B\) is the mid - point of \(AC\), and \(AB = DB\) (because \(DE\) is a perpendicular bisector, so \(AB = AD\)? No, I think I messed up the diagram. Wait, the correct way: The line \(l\) (containing \(D\) and \(E\)) is the perpendicular bisector of \(AC\), so \(AB = BC\) and \(\angle DBC = 90^{\circ}\). So triangle \(DBC\) is right - angled at \(B\), so \(DB=\sqrt{DC^{2}-BC^{2}}=\sqrt{10^{2}-6^{2}}=\sqrt{64}=8\). Since \(l\) is the perpendicular bisector, \(AB = DB = 8\) (because \(A\) and \(D\) are symmetric with respect to the perpendicular bisector? No, wait, \(AC\) is bisected at \(B\), so \(AC = 2\times AB\). And since \(AB = DC\)? No, \(AB = DB\). Since \(DB = 8\), then \(AB = 8\), so \(AC=AB + BC=8 + 8 = 16\)? Wait, but \(BC\) is given as \(6\). Oh! I see my mistake. The markings on \(AC\) are equal, so \(AB = BC\), but in the right - triangle \(DBC\), \(BC = 6\), \(DC = 10\), so \(DB=\sqrt{10^{2}-6^{2}} = 8\). And since \(AB = DB\) (because the triangles \(ABD\) and \(CBD\) are congruent: \(AB = BC\) (wait, no, \(AB = BC\) is the mid - point, but \(DB\) is perpendicular, so \(AD = DC\) (by perpendicular bisector theorem: any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment). So \(AD = DC = 10\)? No, \(DC = 10\), so \(AD = 10\). Then in triangle \(ABD\), which is right - angled at \(B\), \(AD = 10\), \(DB = 8\), so \(AB=\sqrt{AD^{2}-DB^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\). Then \(AC = 2\times AB=12\)? Wait, now I'm confused. Let's start over.

Perpendicular Bisector Theorem: If a line is perpendicular to a segment and bisects it, then any point on the line is equidistant from the endpoints of the segment. So \(D\) is on the perpendicular bisector of \(AC\) (since \(DB\perp AC\) and \(AB = BC\) (from the markings on \(AC\))), so \(AD = DC\). Given \(DC = 10\), so \(AD = 10\). In right - triangle \(DBC\), \(DC = 10\), \(BC = 6\), so \(DB=\sqrt{DC^{2}-BC^{2}}=\sqrt{10^{2}-6^{2}}=\sqrt{64}=8\). In right - triangle \(ABD\), \(AD = 10\), \(DB = 8\), so \(AB=\sqrt{AD^{2}-DB^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{36}=6\). Since \(B\) is the mid - point of \(AC\), \(AC = 2\times AB=2\times6 = 12\)? Wait, no, \(BC = 6\), and \(AB = BC\) (because \(B\) is the mid - point), so \(AC=AB + BC=6 + 6 = 12\). Ah! That makes sense. So first, we use the Pythagorean theorem in \(\triangle DBC\) to find \(DB\), then use the perpendicular bisector theorem to know that \(AD = DC\), then use Pythagorean theorem in \(\triangle ABD\) to find \(AB\), and since \(B\) is the mid - point, \(AC = 2AB\).

Wait, let's re - do the steps correctly:

Step1: Find \(DB\) in \(\triangle DBC\)

\(\triangle DBC\) is right - angled at \(B\). By Pythagorean theorem:
\[ DB=\sqrt{DC^{2}-BC^{2}} \]
Substitute \(DC = 10\) and \(BC = 6\):
\[ DB=\sqrt{10^{2}-6^{2}}=\sqrt{100 - 36}=\sqrt{64}=8 \]

Step2: Use Perpendicular Bisector Theorem

Since \(DB\perp AC\) and \(AB = BC\) (from the markings on \(AC\)), \(D\) lies on the perpendicular bisector of \(AC\). So \(AD = DC = 10\) (Perpendicular Bisector Theorem: any point on the perpendicular bisector of a segment is equidistant from the segment's endpoints).

Step3: Find \(AB\) in \(\triangle ABD\)

\(\triangle ABD\) is right - angled at \(B\). By Pythagorean theorem:
\[ AB=\sqrt{AD^{2}-DB^{2}} \]
Substitute \(AD = 10\) and \(DB = 8\):
\[ AB=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6 \]

Step4: Find \(AC\)

Since \(B\) is the mid - point of \(AC\) (because \(AB = BC\)), \(AC=AB + BC\). But \(AB = BC = 6\)? Wait, no, \(AB = 6\) and \(BC = 6\) (since \(B\) is the mid - point), so \(AC=6 + 6 = 12\).

Answer:

12 units