+291 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.
+291 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 \( 6 \, \text{square units} \).
