In a diagram of an 2 isosceles triangles, where ABC is a big triangle and a ADE, is just inside the top portion of ABC.

Isosceles triangle ABC is similar to a isosceles triangle ADE
what is the length of DE, which is the base part of a isosceles triangle ADE ?
Do you just take one fourth of one of the legs which is 12
so therefore it's 3 on the side of the isosceles triangle ADE leg ?
and the base would be 1/4 of 6 which would be 1.5 or 3/2?

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To find the length of DE, which is the base of the isosceles triangle ADE, we need to use the concept of similarity between the triangles ABC and ADE.

Since triangle ABC is similar to triangle ADE, their corresponding sides are proportional. This means that the ratio of the lengths of their corresponding sides is the same.

In this case, we know that AB is similar to AD, BC is similar to DE, and AC is similar to AE. Let's say the length of AB is 12, and the length of BC is 6.

To find the length of DE, we can set up a proportion using the corresponding sides:

AB/AD = BC/DE

Substituting the given lengths, we have:

12/AD = 6/DE

Now we can solve for DE by cross-multiplying:

6 * AD = 12 * DE

Simplifying the equation further, we get:

AD = 2 * DE

Now, if we substitute AD = 12/4 (1/4 of AB, as mentioned in your question), we can solve for DE:

12/4 = 2 * DE

3 = 2 * DE

Divide both sides by 2 to solve for DE:

DE = 3/2

So the length of DE, which is the base of the isosceles triangle ADE, is 3/2 or 1.5 units.