Jamilla solved the inequality mc016-1. Jpg and graphed the solution as shown below. Mc016-2. Jpg what is the value of b and the missing symbol in jamilla’s inequality? b = –1, mc016-3. Jpg b = –1, mc016-4. Jpg b = 1, mc016-5. Jpg b = 1, mc016-6. Jpg.

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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 space labeled by 2, space labeled by 1 occurs. by corresponding parts of similar triangles, angle bde is congruent to angle bac. 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 space labeled by 2, space labeled by 1 occurs. by corresponding parts of similar triangles, angle bde is congruent to angle bac. 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. δabc ~ δbed 2. side-angle-side similarity postulate 1. δabc ~ δbed 2. side-side-side similarity theorem 1. δabc ~ δdbe 2. side-angle-side similarity postulate 1. δabc ~ δdbe 2. side-side-side similarity theorem