Question

Can anyone help me here asapp,, I am in this question for nearly an hour

Answer 1

Answer:

See below

Step-by-step explanation:

Let side AB equal x. Since triangle ABC is equilateral, sides AB, BC, and Ac are all the same length, x. In any isosceles triangle(equilateral is a type of isosceles triangle) the median is the same as the altitude and angle bisector. This means we can say that AD is also a median. A median splits a side into two equal sections, so we can say BD = DC = x / 2. We are given that DC = CE, so we can also say CE = DC = x / 2. Now, we can use the pythagorean theorem to find the length of AD. So we get the equation:

AB^2 - BD^2 = AD^2

We have the values of AB and BD, so we can substitute them and solve for AD:

x^2 - (x/2)^2 = AD^2

x^2 - x^2 / 4 = AD^2

AD^2 = 3x^2 / 4

AD = x√3 / 2

DE is equal to the sum of DC and CE because of segment addition postulate, so we can say DE = DC + CE = x / 2 + x/ 2 = x. We can again use the pythagorean theorem to find the length of AE:

AD^2 + DE^2 = AE^2

(x√3 / 2)^2 + x^2 = AE^2

3x^2 / 4 + x^2 = AE^2

AE^2 = 7x^2 / 4

AE = x√7 / 2

Now, we know(from before) that AE squared is 7x^2 / 4. We can say EC squared is x^2 / 4 because EC is x / 2 and x / 2 squared is x^2 / 4. We can also notice that AE squared is 7 times EC squared because 7x^2 / 4 = 7 * x^2 / 4

Therefore, we can come to the conclusion AE^2 = 7 EC^2