Given BE = 11, EA = 11, BD = 7, and DC = 7, show that BED ~BAC

To show that two triangles are similar, we need to prove that their corresponding sides are proportional.

In this case, we are given that BE = 11, EA = 11, BD = 7, and DC = 7. We want to prove that the triangle BED is similar to triangle BAC.

To start, let's compare the sides BE and BA. We can see that both sides have the same length, which is 11.

Next, let's compare the sides BD and BC. We know that BD = 7 and DC = 7. Adding these two sides together, we get BD + DC = 7 + 7 = 14. Now, let's compare this sum with the length of side BC. We can see that BC = BD + DC = 14, which is also equal to the sum of BD and DC.

Lastly, let's compare the remaining sides, DE and AC. Since we know that BE = 11 and EA = 11, we can subtract these lengths from their respective sides to find the length of DE and AC. DE = BE - BD = 11 - 7 = 4, and AC = EA - DC = 11 - 7 = 4. We can see that both DE and AC have the same length, 4.

Now, we have shown that BE/BA = BD/BC = DE/AC, which means that the triangles BED and BAC satisfy the corresponding sides proportionality condition. Therefore, we can conclude that BED ~ BAC.