In δabc shown below, bd over ba equals be over bc:
triangle abc with segment de interse...
Mathematics, 08.11.2019 02:31, wizewolf
In δabc shown below, bd over ba equals be over bc:
triangle abc with segment de intersecting sides ab and bc respectively.
the following flowchart proof with missing statements and reasons proves that if a line intersects two sides of a triangle and divides these sides proportionally, the line is parallel to the third side:
top path, by given the ratio of line segments bd to ba is equal to the ratio of line segments be to bc. by side angle side similarity postulate, triangle abc is similar to triangle dbe. by space labeled by 2, space labeled by 1 occurs. by converse of the corresponding angles postulate, line segment de is parallel to line segment ac. bottom path, by reflexive property of quality, angle b is congruent to angle b. by side angle side similarity postulate, triangle abc is similar to triangle dbe. by space labeled by 2, space labeled by 1 occurs. by converse of the corresponding angles postulate, line segment de is parallel to line segment ac.
which reason can be used to fill in the numbered blank space?
1. ∠bde ≅ ∠bac
2. corresponding angles postulate
1. ∠bde ≅ ∠bac
2. corresponding parts of similar triangles
1. ∠bde ≅ ∠bca
2. alternate exterior theorem
1. ∠bde ≅ ∠bca
2. corresponding parts of similar triangles
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