Question

end of semester test
∠abc ≅ ∠cbd

1. △abc - △acd

△abc - △cbd

1. \\( \frac { a c } { a b } = \frac { a d } { a c } \\) and \\( \frac { b c } { a b } = \frac { d b } { b c } \\)
2. \\( a c ^ { 2 } = ( a b ) ( a d ) \\)

\\( b c ^ { 2 } = ( a b ) ( d b ) \\)

1. \\( a c ^ { 2 } + b c ^ { 2 } = ( a b ) ( a d ) + ( a b ) ( d b ) \\)
2. \\( a c ^ { 2 } + b c ^ { 2 } = a b ( a d + d b ) \\)
3. \\( a b = a d + d b \\)
4. \\( a c ^ { 2 } + b c ^ { 2 } = ( a b ) ( a b ) \\)
5. \\( a c ^ { 2 } + b c ^ { 2 } = a b ^ { 2 } \\)

which reason completes the proof?
a. corresponding sides of similar triangles are proportional.
b. corresponding sides of congruent triangles are proportional.
c. corresponding parts of congruent triangles are congruent.
d. corresponding parts of similar triangles are congruent.

Explanation:

• In step 3, it is given that \(\triangle ABC\sim\triangle ACD\) and \(\triangle ABC\sim\triangle CBD\) by AA (Angle - Angle) similarity criteria.
• When two triangles are similar, the ratios of their corresponding side lengths are equal. That is, if \(\triangle ABC\sim\triangle ACD\), then \(\frac{AC}{AB}=\frac{AD}{AC}\) (cross - multiplying gives \(AC^{2}=(AB)(AD)\)), and if \(\triangle ABC\sim\triangle CBD\), then \(\frac{BC}{AB}=\frac{DB}{BC}\) (cross - multiplying gives \(BC^{2}=(AB)(DB)\)).
• The property that allows us to write \(\frac{AC}{AB}=\frac{AD}{AC}\) and \(\frac{BC}{AB}=\frac{DB}{BC}\) is based on the fact that for similar triangles, the corresponding sides are proportional.

Answer:

A. Corresponding sides of similar triangles are proportional.