+743 Properties of Equilateral Triangle:
Since triangle DEF is equilateral, all three sides (DE, DF, and EF) are equal in length. Additionally, all three angles are also equal to 60 degrees.
Right Triangle ACB:
Given that angle ACB = 90 degrees and CD = 5 and CE = 4, we recognize this as a Pythagorean triple (3, 4, 5). Therefore, AC = 3.
Isosceles Triangle DEC:
Since DE = DC + CE and all sides of triangle DEF are equal, triangle DEC is isosceles with DE = 9.
Dropping an Altitude from A:
Draw an altitude from point A to side DE, intersecting DE at point H. Since triangle ACB is a right triangle with a 90-degree angle at C,
line segment AH is also an altitude of triangle DEC, splitting it into two congruent right triangles (ADH and CHE).
Finding AH:
In right triangle ADH, we know the hypotenuse (AD = AC = 3) and want to find the altitude (AH).
Since angle A is a vertex of an equilateral triangle, angle HAD = 60 degrees (half of an equilateral triangle's angle).
We can use the trigonometric ratio tangent (tan) for this situation: tan(HAD) = AH/HD.
Knowing tan(60) = √3 and that HD = DE/2 (since AH bisects DE), we can set up the equation: tan(60) = √3 = AH / (DE/2)
Solving for AH: AH = √3 * (DE/2) = √3 * (9/2) = 3√3 / 2
Finding AE:
In right triangle AEH, we know the altitude (AH = 3√3 / 2) and the base (HE = CE - CH = 4 - AC = 4 - 3 = 1).
We can use the Pythagorean theorem: AE^2 = AH^2 + HE^2
Substitute the known values: AE^2 = (3√3 / 2)^2 + 1^2 = 27/4 + 1 = 31/4
Take the square root of both sides to find AE: AE = sqrt(31)/2
Therefore, the length of AE is equal to sqrt(31)/2.
