Quadrilateral inside Triangle

By the bisector theorem in $\triangle CEB$ and pythagorean theorem (to find $EB$ )

We get that $EM = \frac{3 \sqrt{73}}{26}$ and $MB = \frac {5 \sqrt{73}}{13}$

Again with bisector theorem but in $\triangle ABC$ we get that $AD = \frac{3}{2}$ and $DB = \frac{5}{2}$

Now, by triangle similiarity $\triangle ABE$ ~ $\triangle DHB$ with this we can find $DH$ and then calculate $\triangle DMB$ area (we know $MB$ ) in order to subtract from $\triangle ABE$ .

Doing the boring calculation we get $3 - \frac{75}{52} = \frac{81}{52}$ then answer is $133$

$LaTeX$ bisector theorem gives $AD=\frac{3}{8}\cdot 4 = 1\frac{1}{2}$

$BE:y = -\frac{3}{8}x+1\frac{1}{2}$ and $CD: y = -2x+3$ intersect at $(\frac{12}{13},\frac{15}{13})$

Divide $ADME$ in two triangle and a square

Area $=\frac{1}{2} \cdot(1\frac{1}{2}-\frac{15}{13})\cdot \frac{12}{13} +\frac{12}{13} \cdot \frac{15}{13} + \frac{1}{2} \cdot (1\frac{1}{2} - \frac{12}{13})\cdot \frac{15}{13}=\frac{81}{52}$ so the answer is $81+52=133$

In my solution, I had found the ratios $\frac{AD}{DB},\frac{CM}{MD}$ by using Menelaus' theorem then by using the fact that Cevians divide the sides in the ratio equal to the ratio of the corresponding areas on either sides of the cevian, we will get the areas of $\Delta CBE$ and $\Delta MBD$ . Then by subtracting this area from the total area of the triangle get the answer.

We know $AC=3$ and $AB=4$ , Therefore by Pythagorus theorem , $BC=5$ and the area of right triangle $ACB=\frac{1}{2}\times3\times4=6$ .

Also by Internal Angle Bisector Theorem ,

$\large\frac{AD}{DB}=\frac{AC}{CB}$

$\large\Rightarrow \frac{AD}{DB}=\frac{3}{5}$

Now observe that in $\Delta CAD, EB$ divides the $3$ sides of triangle in ratios $\frac{CE}{EA},\frac{AB}{BD},\frac{BM}{MC}$

Therefore, by Menelaus's theorem

$\large\frac{CE}{EA}\times\frac{AB}{BD}\times\frac{BM}{MC}=1$ and we know $\large\frac{CE}{EA}=1$

Putting the known ratios,

$\large\frac{1}{1}\times\frac{8}{5}\times\frac{BM}{MC}=1$ $\large\Rightarrow \frac{BM}{MC}=\frac{5}{8}$

So now area of

$\large\Delta MBD=\frac{MD}{DC}\times\frac{DB}{AB}\times\text{ Area of }\Delta ABC=\frac{150}{104}$

and the area of $\large\Delta CBE=\frac{1}{2}\times\text{ Area of }\Delta ABC=3$ .

So the area of quadrilateral $\large AEMD=6-3-\frac{150}{104}=\boxed{\frac{81}{52}}$

So the answer is $\boxed{133}$

Place the triangle on a coordinate plane so that $A$ is at $(0, 0)$ . Then $B$ is at $(4, 0)$ , $C$ is at $(0, 3)$ , and $E$ is at $(0, \frac{3}{2})$ .

Since $CD$ is the angle bisector of $\angle ACB$ with $AC = 3$ and $AB = 4$ , $AD = 3 \tan (\frac{1}{2} \tan^{-1} (\frac{4}{3}))$ . Using the tangent half angle formula and Pythagorean's Theorem, $AD = 3 \cdot \frac{1 - \frac{3}{5}}{\frac{4}{5}} = \frac{3}{2}$ . Therefore, $D$ is at $(\frac{3}{2}, 0)$ .

Using the above coordinates, $CD$ is on line $y = -2x + 3$ and $EB$ is on line $y = -\frac{3}{8}x + \frac{3}{2}$ . The intersection $M$ is therefore at the solution to the two lines which is at $(\frac{12}{13}, \frac{15}{13})$ .

This means that $\triangle AMD$ has a base of $\frac{3}{2}$ and a height of $\frac{15}{13}$ , and so an area of $\frac{1}{2}\frac{3}{2}\frac{15}{13} = \frac{45}{52}$ ; and $\triangle AME$ has a base of $\frac{3}{2}$ and a height of $\frac{12}{13}$ , and so an area of $\frac{1}{2}\frac{3}{2}\frac{12}{13} = \frac{36}{52}$ .

The area of quadrilateral $ADME$ is equal to the sum of the areas of $\triangle AMD$ and $\triangle AME$ , which is $\frac{45}{52} + \frac{36}{52} = \frac{81}{52}$ , and so $a = 81$ and $b = 52$ , and $a + b = \boxed{133}$ .

Use coordinate Geometry, with A(0,0). And <ACD= ½ <ACB = ½ sin-1(4/5) = 26.625. So AD=3tan 26.625 = 3/2.We have D(3/2,0), E(0,3/2). The eqs. ofline CD : y = -2x + 3, and BE: y=(-3/8) x +3/2. The intersection point M(12/13, 15/13). The area of quadrilateral ADME= the area of ΔADM + the area of ΔADM + ΔAEM = ½ (3/2 . 15/13 + 3/2 . 12/13 ) = 81/52. So a+b=133