Your teacher will grade your response for questions 13-15, to ensure that you receive proper credit for your answers write a paragraph proof.
Given : BC = EC and AC = DC
Prove : BA = ED
To prove that BA = ED under the given conditions, we begin by examining the given information. We have a common structure, likely a geometric figure, where points A, B, C, D, and E are all connected in some manner, and we know that BC = EC and AC = DC. Even though the precise geometric figure is not described in this scenario, these conditions imply that there are two isosceles triangles sharing a common base, namely triangle ABC and triangle ADC, since they both have two sides of equal length.
In triangle ABC, the sides are such that AC is equal to BC. In triangle ADC, we have AC equal to DC. Given that both triangles share side AC, and it is equal in length to BC in one triangle and equal in length to DC in another triangle, we can conclude that BC is also equal to DC.
Now that we have established that BC is equal to DC, we can observe the entire figure. Triangle ABC and triangle ADC share a common vertex at point C. Since BC and DC are equal as proven, BCD is an isosceles triangle. Also, since BC and EC are equal by the given, BCE is also an isosceles triangle. Given congruent segments BC and EC, angle B must equal angle E by the Isosceles Triangle Theorem (base angles in an isosceles triangle are equal).
As angles B and E are equal, and we have established that BC = EC and DC = BC, triangle BCD is essentially "mirrored" to create triangle EDC. Because BCD is isosceles with two equal sides BC and DC, and EDC is the mirror image, it becomes evident that BA must be equal to ED, since the corresponding sides in congruent triangles or mirrored isosceles triangles are also equal.
Therefore, by using the properties of isosceles triangles and the given equalities, we have sufficiently shown that BA must be equal to ED.