Geometry help!

2) In a certain iscoceles right triangle, the altitude to the hypotenuse has length 4sqrt2. What is the area of the triangle?

Area  =  (1/2)hypotenuse * (altidude drawn to the hypotenuse)   (1)

Area  (1/2)  product of the leg lengths    (2)

Equate (1) and (2)  and let the leg lengths  = L  and we have

(1/2) hypotenuse * ( 4sqrt 2 )  = (1/2) L^2     multiply through by 1/2

hypotenuse * ( 4sqrt(2)   =  L^2

The hypotenuse  =  sqrt  ( L^2 + L^2)  =  sqrt (2 L^2 )  =   Lsqrt 2

L sqrt 2 * 4* sqrt 2  = L^2    divide through  by L

4 sqrt 2 * sqrt 2  = L

4*2  = L

8 = L

So...the area  =  (1/2) L^2  = (1/2) * 8^2  =  32 units^2

1) In triangle ABC, AB=AC and D is a point on line AC so that line BD bisects angle ABC. If BD=BC, what is the measure, in degrees, of angle A?

Call the each bisected angle. a

And call angle BCA, b

And since BD  = BC...then angle BDC  = angle BCD  = angle BCA  = b

So....since AB = AC....then angle ABC  = angle BCA ⇒  2a  = b ⇒  4a  = 2b

So..in triangle BDC....

a + 2b  =180

a + 4a  = 180

5a  = 180

a  = 36

And  b  = 2a =  72

So...in triangle ABC

Angle ABC  + Angle BCA  + Angle A  = 180

2a + b   +  Angle A  = 180

72  +  72  + Angle A  =  180

144  + Angle A  =  180

Angle A  =  180  - 144  =  108°

3) In triangle ABC, AB=17, AC=*, and BC=15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

If AC was supposed to be 8, we have a right triangle....

Its area  = (1/2) product of the legs  = (1/2) AC * BC  =  (1/2) 8 * 15  =  60  units^2