PLEASE HELP

The diagram represents a residential subdivision in the shape of a right triangle. A road from point C to point D bisects the right angle C. How far is the intersection at
point D from the corner of the subdivision at point A, to the nearest tenth of a meter?
300 m
The intersection at point D is 400 meters from the comer of the subdivision at point A.

DA = 285.7 m

Step-by-step explanation:

First we need to find the side AB in the triangle ABC, and we can do this using Pythagoras' theorem:

AB^2 = BC^2 + AC^2

AB^2 = 300^2 + 400^2

AB^2 = 25000

AB = 500 m

We can find the angle ABC with the tangent relation:

tangent(ABC) = 400/300 = 4/3

ABC = 53.13°

From triangle ABC, we have:

ABC + BCA + CAB = 180°

53.13 + 90 + CAB = 180

CAB = 36.87°

From triangle DAC, we have:

DAC + ACD + CDA = 180

36.87 + 45 + CDA = 180

CDA = 98.13°

Now to find the side of DA, we can use law of sines in triangle DAC:

DA/sin(DCA) = AC/sin(CDA)

DA/sin(45) = 400/sin(98.13)

DA = 400 * 0.7071 / 0.9899 = 285.7258 m

Rounding to nearest tenth, we have DA = 285.7 m