In the figure, ABCD is a square and ABE is an equilateral triangle. Find <ADE. I just for the life of me find out what E and D are... I got A, but... Grrrr...
AB = AE because triangle AEB is equilateral.
AD = AB because ABCD is a square.
AD = AE because AD = AB = AE.
Thus triangle EAD is isosceles.
Thus angle ADE = angle AED.
Angle DAE is the complement of angle EAB because the interior angles of a square are right angles.
Angle EAB is 60 degrees because triangle AEB is equilateral.
Thus angle DAE is 90 - 60 = 30 degrees.
Call angle AED x. Then 30 + x + x = 180 because the sum of the interior angles of a triangle is 180 degrees.
Solving this equation I get that x = 75, so angle AED = angle ADE= 75 degrees.
-Dan
This really should have gone in its own thread. See rule #15 here.
$\displaystyle (x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4)$
Expanding
$\displaystyle x^2 - 6x + 8 = x^2 - 8x + 15 - Ax + 3A - Bx + 4B$
Collecting like terms:
$\displaystyle (-6 + 8 + A + B)x + (8 - 15 - 3A - 4B) = 0$
$\displaystyle (A + B + 2)x + (-3A - 4B - 7) = 0$
In order for this expression to be 0 for all x we must have the coefficients of all powers of x equal to 0. So
$\displaystyle A + B + 2 = 0$
and
$\displaystyle -3A - 4B - 7 = 0$
Can you take things from here?
-Dan