MHB Find Length of Line D in ABC Triangle

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To find the length of line segment D in triangle ABC, where AB = 3, AC = 4, and BC = 6, the ratio BD:DC is given as 2:3. By defining u and v for segments BD and DC respectively, the equations u + v = 6 and u/v = 2/3 lead to the calculation of v as 18/5. The Law of Cosines is then applied to find the cosine of angle θ, resulting in cos(θ) = 43/48. Finally, using the Law of Cosines again, the length of segment D is calculated as x = √(4^2 + (18/5)^2 - 2 * 4 * (18/5) * (43/48)), yielding x = √79/5.
MarkFL
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Here is the question:

How will you find the length of a line drawn inside a scalene triangle?


ABC is a scalene triangle, in which AB = 3, AC = 4 and BC = 6
D is the line joining A with BC.such that BD : DC : :2 :3

I have posted a link there to this topic so the OP can see my work.
 
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Re: Sepia's question ay Yahoo! Answers regarding finding the length of a line segment within a trian

Hello Sepia,

Let's first draw a diagram:

View attachment 1408

We see that:

$$u+v=6$$

and we are given:

$$\frac{u}{v}=\frac{2}{3}\implies u=\frac{2}{3}v$$

Substituting this into the first equation, we then find:

$$\frac{2}{3}v+v=6$$

$$\frac{5}{3}v=6$$

$$v=\frac{18}{5}$$

Now, let's use the Law of Cosines to determine the cosine of the angle $\theta$:

$$3^2=4^2+6^2-2\cdot4\cdot6\cos(\theta)$$

$$\cos(\theta)=\frac{4^2+6^2-3^2}{2\cdot4\cdot6}=\frac{43}{48}$$

Next, using the Law of Cosines again, we may state:

$$x=\sqrt{4^2+v^2-2\cdot4\cdot v\cos(\theta)}$$

$$x=\sqrt{4^2+\left(\frac{18}{5} \right)^2-2\cdot4\cdot\left(\frac{18}{5} \right)\left(\frac{43}{48} \right)}=\frac{\sqrt{79}}{5}$$
 

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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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