Hypotenuse middle point

Segment $AD$ will have length $5$ , half the hypotenuse of $\triangle ABC$ . (The easiest way to see this is to consider the circumcircle of $\triangle ABC$ ; $BC$ must be a diameter as it subtends a right angle, thus $AD$ is a radius.)

So $\triangle ADB$ is isosceles. If we drop a perpendicular segment from $D$ meeting $AB$ at $F$ , then $F$ bisects $AB$ , so $AF=3$ , and $\triangle ADF$ is $3 \text{-} 4 \text{-} 5$ . Also, $\triangle ADF \sim \triangle EAD$ , as $\angle FAD$ and $\angle DAE$ are complementary. Then

$\dfrac{AF}{DE} = \dfrac{DF}{AD} \implies \dfrac{3}{DE} = \dfrac{4}{5} \implies DE = \dfrac{15}{4}$

so the area of $\triangle ADE$ is $\dfrac{1}{2} \cdot 5 \cdot \dfrac{15}{4} = \boxed{\dfrac{75}{8}}$

From $\triangle ABC$ , $\angle B = \cos^{-1} \frac{3}{5}$ . Then using the law of cosines on $\triangle ABD, AD^2 = BD^2 + AB^2 - 2 \cdot BD \cdot AB \cdot \cos B$ or $AD^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cdot \cos (\cos^{-1} \frac{3}{5})$ , which solves to $AD = 5$ .

$\triangle ABD$ is an isosceles triangle, so $\angle BAD = \angle B = \cos^{-1} \frac{3}{5}$ . Then $\angle DAE = \angle BAE - \angle BAD = 90° - \cos^{-1} \frac{3}{5}$ , and since $\triangle DEA$ is a right triangle, $\angle DEA = 90° - \angle DAE$ . Combining these gives $\angle DEA = 90° - (90° - \cos^{-1} \frac{3}{5}) = \cos^{-1} \frac{3}{5}$ .

Since $\angle B = \angle DEA = \cos^{-1} \frac{3}{5}$ and $\angle BAC = \angle ADE = 90°$ , $\triangle DEA ~ \triangle ABC$ by AA similarity, and $\frac{DE}{AD} = \frac{AB}{AC}$ or $\frac{DE}{5} = \frac{6}{8}$ , which means $DE = \frac{15}{4}$ .

The area of $\triangle ADE$ is $\frac{1}{2} \cdot AD \cdot DE = \frac{1}{2} \cdot 5 \cdot \frac{15}{4} = \boxed{\frac{75}{8}}$ .

The green triangle is in the same proportions as the $6, 8, 10$ triangle, but scaled down by a factor of $\dfrac{5}{8}$ , where $5$ is half of the hypotenuse, and $8$ is the height of the original triangle. Hence, the area of the green triangle is

$(\dfrac{5}{8})^2 \cdot \dfrac{1}{2}\cdot 6\cdot8=\dfrac{75}{8}$

I have written my solution down on paper

Let the X and Y axis be along AB and AC respectively........then its simple coordinate geometry.......!!