1. ## Angles... :(

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...

2. And then another question... I don't get where my teacher gets this stuff...

(x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4) is an identity in x, find the values of A and B.

Any and all help is appreciated!

3. Originally Posted by Rocher
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.

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

4. Originally Posted by Rocher
And then another question... I don't get where my teacher gets this stuff...

(x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4) is an identity in x, find the values of A and B.

Any and all help is appreciated!
This really should have gone in its own thread. See rule #15 here.

$(x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4)$

Expanding
$x^2 - 6x + 8 = x^2 - 8x + 15 - Ax + 3A - Bx + 4B$

Collecting like terms:
$(-6 + 8 + A + B)x + (8 - 15 - 3A - 4B) = 0$

$(A + B + 2)x + (-3A - 4B - 7) = 0$

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
$A + B + 2 = 0$
and
$-3A - 4B - 7 = 0$

Can you take things from here?

-Dan