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# Geometry

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Let ABC be a triangle, and let its angle bisectors be AD, BE, and CF which intersect at I. If DI=3, BD=4 and BI=6 then compute the area of triangle BID.

Apr 28, 2024

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To find the area of triangle BID, we can use the formula for the area of a triangle given the length of one side and the lengths of the two adjacent sides to an angle. The formula is:

$\text{Area} = \frac{1}{2} \times \text{side} \times \text{adjacent side} \times \sin(\text{angle})$

Given:
- $$DI = 3$$
- $$BD = 4$$
- $$BI = 6$$

We need to find the angle at vertex B.

We know that the angle bisectors of a triangle divide the opposite side into segments that are proportional to the adjacent sides. Therefore, from the given information, we can set up the following proportions:

$\frac{DI}{BD} = \frac{DI}{DI + IB} = \frac{3}{4}$

$\frac{3}{4} = \frac{3}{3 + IB}$

$3 + IB = 4$

$IB = 1$

Now, we can find $$\sin(\angle B)$$ using the Law of Sines in triangle BDI:

$\sin(\angle B) = \frac{DI}{BI} = \frac{3}{6} = \frac{1}{2}$

Now, we can use the formula for the area of triangle BID:

$\text{Area} = \frac{1}{2} \times BD \times DI \times \sin(\angle B)$

$\text{Area} = \frac{1}{2} \times 4 \times 3 \times \frac{1}{2}$

$\text{Area} = 6 \, \text{square units}$

So, the area of triangle BID is $$6 \, \text{square units}$$.

Apr 28, 2024