To predict the length of AC, we notice that △ ABC and △ DBE are similar triangles. This is because they have three pairs of congruent angles. The top angle, ∠ B, is shared by the triangles. According to the Reflexive Property of Congruence we know this angle is congruent in our triangles.
Since DE is the midsegment of AB and BC, it divides these sides in two equal halves. This must mean the sides of △ DBE are half that of △ ABC. Therefore, AC should be twice that of DE.
2DE ⇒ 2( 5)=10 units.
c Like in Part B, we have to use the Distance Formula once more. This time, we will substitute points A(-1, -1) and C(7, 5) for (x_1, y_1) and (x_2, y_2), respectively.
d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
⇕
d = sqrt(( 7 -( - 1))^2 + ( 5 - ( - 1))^2)
Then, let's simplify the expression on the right-hand side.
d Consider the following situation. We have a segment XZ and its midpoint Y. Now, let's also consider the midpoint W, midpoint of XW. Then Y is halfway from X to Z and W is halfway from X to Y. Therefore, W is 12 * 12 = 14 of the way from X to Z.
We want to find points F and G, which are 14 of the way from point B to points A and C, respectively. Based on the above reasoning, these points are midpoints of segments BD and BE, respectively.
Let's find these points using the Midpoint Formula, starting with F.
Therefore, the coordinates of point G are (2.5, 8). Notice that F G is a midsegment of △_(BDE).
Thus, by the Triangle Midsegment Theorem its length is half the length of DE.
FG = DE/2
Recall that from Part B we know that AC is twice as long as DE, or, equivalently, DE is half as long as AC.
DE = AC/2
Combining the two relations, we get that FG is 14 as long as AC.
FG = DE/2 = AC/4
