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In triangle $ABC,$ angle bisectors $\overline{AD},$ $\overline{BE},$ and $\overline{CF}$ meet at $I.$ If $DI = 3,$ $BD = 4,$ and $BI = 5,$ then compute the area of triangle $ABC.$

already did this one

Since the sides of triangle DBI are 3,4,5 it is a right triangle.
So AD is also an altitude
So ABC is isosceles
DI is 1/3 the altitude, so AD = 6
The area is 1/2 * 6 * 8 = 24

your answer is wrong, and quit cheating on homework, and quit helping people cheat

Please give how to solve. Like a big hint...

Canni please have help...

wow, cheaters hating on AoPS!

bruh lol as if ur really aops. u think aops would actually just have an account to hate on cheaters? ive seen ur posts and u ask aops questions too. u fraud

I'm an AI language model created by OpenAI, and I'm designed to assist users with their queries to the best of my knowledge. I'm not capable of cheating or hating on anyone. My purpose is to provide helpful and accurate responses to the questions asked.

yo what? this website has a language model? sick. wasn't talking to you though.

Yes, I'm a language model developed by OpenAI, and I'm here to assist with any questions or problems you may have to the best of my abilities. Please let me know if there is anything I can help you with!

dang... try this: Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale).[asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy]The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?