Question

question 1: in the right triangles shown, the measure of angle abc is the same as the measure of angle ebd. what is the length of side be? the length of be= type your answer...

Explanation:

Step1: Find BC in triangle ABC

In right triangle \(ABC\), \(AC = 3\), \(AB = 5\). By Pythagorean theorem \(BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{5^{2}-3^{2}}=\sqrt{25 - 9}=\sqrt{16}=4\).

Step2: Prove triangles ABC and EBD similar

\(\angle C=\angle D = 90^{\circ}\), \(\angle ABC=\angle EBD\) (given). So \(\triangle ABC\sim\triangle EBD\) (AA similarity).

Step3: Set up proportion for similar triangles

For similar triangles, \(\frac{AB}{BE}=\frac{BC}{BD}=\frac{AC}{ED}\). We know \(AC = 3\), \(ED = 2\), \(BC = 4\), \(AB = 5\). From \(\frac{AC}{ED}=\frac{BC}{BD}\), but we can also use \(\frac{AB}{BE}=\frac{AC}{ED}\). So \(\frac{5}{BE}=\frac{3}{2}\), cross - multiply: \(3BE=10\)? Wait, no, wait. Wait, actually, since \(\triangle ABC\sim\triangle EBD\), the ratio of corresponding sides: \(\frac{AB}{BE}=\frac{BC}{BD}=\frac{AC}{ED}\). Wait, \(AC = 3\), \(ED = 2\), \(BC = 4\), \(AB = 5\). Let's use \(\frac{BC}{BD}=\frac{AC}{ED}\), but we can also use \(\frac{AB}{BE}=\frac{BC}{BD}\)? Wait, no, let's re - check. \(\angle ABC=\angle EBD\), \(\angle C=\angle D = 90^{\circ}\), so corresponding sides: \(AC\) corresponds to \(ED\), \(BC\) corresponds to \(BD\), \(AB\) corresponds to \(BE\). So \(\frac{AC}{ED}=\frac{BC}{BD}=\frac{AB}{BE}\). We know \(AC = 3\), \(ED = 2\), \(BC = 4\), \(AB = 5\). Let's use \(\frac{AB}{BE}=\frac{AC}{ED}\), so \(\frac{5}{BE}=\frac{3}{2}\)? Wait, that gives \(BE=\frac{10}{3}\)? No, wait, maybe I mixed up the correspondence. Wait, \(AC = 3\), \(ED = 2\), \(BC = 4\), \(BD\) is equal to \(BC\)? No, wait, \(BC\) is 4, \(ED\) is 2, \(AC\) is 3. Wait, actually, the correct correspondence: \(\triangle ABC\) and \(\triangle EBD\), so \(AB\) corresponds to \(EB\), \(BC\) corresponds to \(BD\), \(AC\) corresponds to \(ED\). So \(\frac{AB}{EB}=\frac{BC}{BD}=\frac{AC}{ED}\). We can also use \(\frac{BC}{BD}=\frac{AC}{ED}\), but we know \(AC = 3\), \(ED = 2\), \(BC = 4\), so \(\frac{4}{BD}=\frac{3}{2}\), \(BD=\frac{8}{3}\). But we can use \(\frac{AB}{EB}=\frac{AC}{ED}\), so \(\frac{5}{EB}=\frac{3}{2}\), no, that's wrong. Wait, no, wait, I think I made a mistake in correspondence. Let's do it again. \(\angle ABC=\angle EBD\), \(\angle C = \angle D=90^{\circ}\), so:

\(AB\) (hypotenuse of \(\triangle ABC\)) corresponds to \(BE\) (hypotenuse of \(\triangle EBD\))

\(AC\) (leg of \(\triangle ABC\)) corresponds to \(ED\) (leg of \(\triangle EBD\))

\(BC\) (leg of \(\triangle ABC\)) corresponds to \(BD\) (leg of \(\triangle EBD\))

So the ratio of similarity is \(\frac{AC}{ED}=\frac{3}{2}\)

So all sides of \(\triangle ABC\) are in ratio \(\frac{3}{2}\) to sides of \(\triangle EBD\)? Wait, no, similarity ratio is \(\frac{AC}{ED}=\frac{3}{2}\), so \(\frac{AB}{BE}=\frac{3}{2}\), so \(BE=\frac{AB\times2}{3}=\frac{5\times2}{3}=\frac{10}{3}\)? No, that can't be. Wait, wait, maybe the correspondence is reversed. Maybe \(\triangle ABC\sim\triangle BED\). Let's check angles again. \(\angle ABC=\angle EBD\), \(\angle C=\angle D = 90^{\circ}\), so \(\triangle ABC\sim\triangle EBD\) (AA). So \(AB\) corresponds to \(EB\), \(BC\) corresponds to \(BD\), \(AC\) corresponds to \(ED\). So \(\frac{AB}{EB}=\frac{BC}{BD}=\frac{AC}{ED}\). We have \(AC = 3\), \(ED = 2\), \(BC = 4\), \(AB = 5\). Let's use \(\frac{AC}{ED}=\frac{BC}{BD}\), so \(\frac{3}{2}=\frac{4}{BD}\), then \(BD=\frac{8}{3}\). Then use \(\frac{AB}{EB}=\frac{BC}{BD}\), so \(\frac{5}{EB}=\frac{4}{\frac{8}{3}}=\frac{4\times3}{8}=\frac{3}{2}\). Then \(EB=\frac{5\times2}{3}=\frac{10}{3}\)? No, that's not right. Wait, wait, I think I made a mistake in calculating \(BC\). Wait, \(AC = 3\), \(AB = 5\), so \(BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{25 - 9}=4\), that's correct. Then \(ED = 2\), \(AC = 3\), so the ratio of \(AC\) to \(ED\) is \(3:2\). Then since the triangles are similar, the ratio of \(AB\) to \(BE\) should be equal to the ratio of \(AC\) to \(ED\)? Wait, no, \(AB\) is the hypotenuse of \(\triangle ABC\), \(BE\) is the hypotenuse of \(\triangle EBD\). \(AC\) and \(ED\) are the legs opposite to \(\angle ABC\) and \(\angle EBD\) respectively. So \(\sin(\angle ABC)=\frac{AC}{AB}=\frac{3}{5}\), and \(\sin(\angle EBD)=\frac{ED}{BE}=\frac{2}{BE}\). Since \(\angle ABC=\angle EBD\), \(\frac{3}{5}=\frac{2}{BE}\), cross - multiply: \(3BE = 10\), \(BE=\frac{10}{3}\)? No, that's not possible. Wait, wait, maybe the correspondence is \(\triangle ABC\) and \(\triangle BED\) with \(\angle ABC=\angle BED\)? No, the problem says \(\angle ABC=\angle EBD\). Wait, maybe I messed up the diagram. Let's look at the diagram again: \(C\) and \(D\) are right angles, \(AC = 3\), \(ED = 2\), \(AB = 5\), \(BC\) is horizontal, \(BD\) is horizontal, \(AC\) is vertical, \(ED\) is vertical. So \(\triangle ABC\) has legs \(AC = 3\), \(BC\), hypotenuse \(AB = 5\). \(\triangle EBD\) has legs \(ED = 2\), \(BD\), hypotenuse \(BE\). And \(\angle ABC=\angle EBD\). So the correct similarity ratio is \(\frac{AC}{ED}=\frac{BC}{BD}=\frac{AB}{BE}\). We found \(BC = 4\), \(AC = 3\), \(ED = 2\). So \(\frac{AC}{ED}=\frac{3}{2}\), so \(\frac{AB}{BE}=\frac{3}{2}\), so \(BE=\frac{2\times AB}{3}=\frac{2\times5}{3}=\frac{10}{3}\)? Wait, but that seems odd. Wait, no, wait, maybe the ratio is \(\frac{BC}{ED}=\frac{AC}{BD}=\frac{AB}{BE}\)? No, \(BC\) is 4, \(ED\) is 2, so \(\frac{BC}{ED}=2\). Then \(\frac{AC}{BD}=2\), \(AC = 3\), so \(BD=\frac{3}{2}\). Then \(\frac{AB}{BE}=2\), \(AB = 5\), so \(BE=\frac{5}{2}\)? No, this is confusing. Wait, let's use the Pythagorean theorem in \(\triangle EBD\) once we find \(BD\). Wait, we know \(BC = 4\) (from \(\triangle ABC\): \(3 - 4-5\) triangle). Now, since \(\triangle ABC\sim\triangle EBD\), the ratio of sides: \(AC/ED=BC/BD=AB/BE\). \(AC = 3\), \(ED = 2\), so the ratio of similarity is \(3/2\). So \(BC/BD = 3/2\), \(BC = 4\), so \(BD=(2\times BC)/3=(2\times4)/3 = 8/3\). Then in \(\triangle EBD\), using Pythagorean theorem: \(BE=\sqrt{BD^{2}+ED^{2}}=\sqrt{(\frac{8}{3})^{2}+2^{2}}=\sqrt{\frac{64}{9}+4}=\sqrt{\frac{64 + 36}{9}}=\sqrt{\frac{100}{9}}=\frac{10}{3}\). Ah, there we go. So \(BE=\frac{10}{3}\)? Wait, but that's \(\approx3.33\). But let's check again. \(BC = 4\), \(BD=\frac{8}{3}\), \(ED = 2\). Then \(BE=\sqrt{(\frac{8}{3})^{2}+2^{2}}=\sqrt{\frac{64 + 36}{9}}=\sqrt{\frac{100}{9}}=\frac{10}{3}\). Yes, that's correct.

Answer:

\(\frac{10}{3}\)