Question

proof create a paragraph proof of the triangle proportionality theorem (theorem 8.8) by dragging the statements into the correct order. given: $overline{bd} \parallel overline{ae}$ prove: $\frac{ab}{cb} = \frac{de}{cd}$ proof: keyboard help

Explanation:

Step1: Identify Similar Triangles

Since \( \overline{BD} \parallel \overline{AE} \), by the Basic Proportionality Theorem (Thales' theorem) or the AA (Angle - Angle) similarity criterion, \( \triangle CBD \sim \triangle CAE \). The corresponding angles are equal: \( \angle 1=\angle 4 \) (corresponding angles, as \( BD\parallel AE \)) and \( \angle 2 = \angle 3 \) (corresponding angles, as \( BD\parallel AE \)), and \( \angle C \) is common to both triangles.

Step2: Use Proportionality of Similar Triangles

For similar triangles \( \triangle CBD \) and \( \triangle CAE \), the ratios of corresponding sides are equal. That is, \( \frac{CB}{CA}=\frac{CD}{CE} \). We can rewrite \( CA=CB + BA \) and \( CE=CD + DE \). So, \( \frac{CB}{CB + BA}=\frac{CD}{CD + DE} \). Cross - multiplying gives \( CB(CD + DE)=CD(CB + BA) \). Expanding both sides: \( CB\cdot CD+CB\cdot DE = CD\cdot CB+CD\cdot BA \). Subtracting \( CB\cdot CD \) from both sides, we get \( CB\cdot DE=CD\cdot BA \). Rearranging the terms, we have \( \frac{BA}{CD}=\frac{DE}{CB} \)? Wait, no, let's start over with the correct proportion. Since \( \triangle CBD\sim\triangle CAE \), \( \frac{CB}{CA}=\frac{CD}{CE} \), and \( CA = CB + AB \), \( CE=CD + DE \). So \( \frac{CB}{CB + AB}=\frac{CD}{CD + DE} \). Cross - multiply: \( CB(CD + DE)=CD(CB + AB) \). \( CB\cdot CD+CB\cdot DE=CD\cdot CB + CD\cdot AB \). Subtract \( CB\cdot CD \) from both sides: \( CB\cdot DE = CD\cdot AB \). Then \( \frac{AB}{CD}=\frac{DE}{CB} \)? No, the correct proportion from the similarity of \( \triangle CBD \) and \( \triangle CAE \) is \( \frac{AB}{CB}=\frac{DE}{CD} \)? Wait, maybe a better approach. Since \( BD\parallel AE \), by the Basic Proportionality Theorem (Thales' theorem), which states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. In \( \triangle CAE \), line \( BD \) is parallel to \( AE \) and intersects \( CA \) at \( B \) and \( CE \) at \( D \). So, \( \frac{AB}{CB}=\frac{DE}{CD} \), which can be rewritten as \( \frac{AB}{CD}=\frac{DE}{CB} \)? No, the correct statement from Thales' theorem (Triangle Proportionality Theorem) is that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the two sides proportionally. So in \( \triangle CAE \), with \( BD\parallel AE \), we have \( \frac{AB}{BC}=\frac{DE}{CD} \), which is equivalent to \( \frac{AB}{CD}=\frac{DE}{BC} \)? Wait, the problem wants to prove \( \frac{AB}{CD}=\frac{DE}{CB} \)? Wait, the given to prove is \( \frac{BA}{CD}=\frac{DE}{CB} \) (assuming \( BA = AB \) and \( CB \) is as in the diagram). Wait, maybe the initial identification of similar triangles is correct. Since \( BD\parallel AE \), \( \angle C \) is common, \( \angle CBD=\angle CAE \) (corresponding angles) and \( \angle CDB=\angle CEA \) (corresponding angles). So \( \triangle CBD\sim\triangle CAE \) by AA similarity. Then, the ratio of corresponding sides: \( \frac{CB}{CA}=\frac{CD}{CE} \). Let \( CA=CB + AB \) and \( CE = CD+DE \). So \( \frac{CB}{CB + AB}=\frac{CD}{CD + DE} \). Cross - multiply: \( CB(CD + DE)=CD(CB + AB) \). \( CB\cdot CD+CB\cdot DE=CD\cdot CB + CD\cdot AB \). Subtract \( CB\cdot CD \) from both sides: \( CB\cdot DE = CD\cdot AB \). Then, \( \frac{AB}{CD}=\frac{DE}{CB} \), which is the required result.

Answer:

To prove the Triangle Proportionality Theorem (Thales' theorem) for \( \overline{BD}\parallel\overline{AE} \) in \( \triangle CAE \):

1. Identify Similar Triangles: Since \( \overline{BD}\parallel\overline{AE} \), \( \angle C \) is common, \( \angle CBD=\angle CAE \) (corresponding angles), and \( \angle CDB=\angle CEA \) (corresponding angles). By the AA (Angle - Angle) similarity criterion, \( \triangle CBD\sim\triangle CAE \).

1. Use Proportionality of Similar Triangles: For similar triangles \( \triangle CBD \) and \( \triangle CAE \), the ratios of corresponding sides are equal. So, \( \frac{CB}{CA}=\frac{CD}{CE} \). Let \( CA = CB + AB \) and \( CE=CD + DE \). Substituting these into the proportion gives \( \frac{CB}{CB + AB}=\frac{CD}{CD + DE} \).

1. Cross - Multiply and Simplify: Cross - multiplying the proportion \( \frac{CB}{CB + AB}=\frac{CD}{CD + DE} \) gives \( CB(CD + DE)=CD(CB + AB) \). Expanding both sides: \( CB\cdot CD+CB\cdot DE = CD\cdot CB+CD\cdot AB \). Subtracting \( CB\cdot CD \) from both sides, we obtain \( CB\cdot DE = CD\cdot AB \). Rearranging the terms, we get \( \frac{AB}{CD}=\frac{DE}{CB} \) (or \( \frac{BA}{CD}=\frac{DE}{CB} \) as required), thus proving the Triangle Proportionality Theorem.