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Pearson Geometry Common Core, 2011 View details

Cumulative Standards Review

Exercise 22 Page 486

Practice makes perfect

a It is given that certain segments have the same lengths, so they are congruent. Let's mark these on the diagram. It is given that BC=AD, so the distance of C from B is the same as the distance of D from E.

However, there is no other constraints for the position of C and D, so the position of these points is not fixed relative to the other points. You can confirm this by moving points C and D on the diagram. Triangles △ BCG and △ EDG are not necessarily congruent.

b Corresponding sides of congruent triangles are congruent. According to the Side-Side-Side (SSS) Congruence Postulate, the converse is also true.

Segment CD is a common side of triangles △ BCD and △ EDC and it is given that BC=ED. If in addition we assume that BD= CE, then triangles △ BCD and △ EDC have three pairs of congruent sides, so they are congruent.

Alternative Solution

Other possibilities

Without detailed explanation, we give here two other pieces of information that also guarantee that triangles △ BCD and △ EDC are congruent.

Piece of Information	Justification
∠ BCD≅ ∠ EDC	Side-Angle-Side (SAS) Congruence Postulate
AC=FD and FC=AD	Along with the given AB=FE, BC=ED, and AE=FB, these guarantee that the diagram is symmetric about the perpendicular bisector of AF.

There are also other possibilities.

c It is given that AB=FE and AE=FB. Since segment AF is a common side of triangles △ BAF and △ EFA, this means that these two triangles have three pairs of congruent sides.

d Since three sides of triangle △ BAF are congruent to three sides of triangle △ EFA, we can use the Side-Side-Side (SSS) Congruence Postulate to conclude that these are congruent triangles.

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Exercises p.484-486

Exercise 22 Page 486

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Alternative Solution