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# Bisectors

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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.

Feb 7, 2024

#1
+394
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Here is a solution that doesn't require any complex geometry. We see that we know all 3 side lengths of triangle BID. We know a formula that can tell us the area directly knowing 3 sides of the triangle, Heron's formula.

First we calculte the semi-perimeter $$\frac{3+4+6}{2}=\frac{13}{2}$$

Then we apply Heron's formula. $$Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}$$. Simplifying, we get $$Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}$$

So, the Area is sqrt(455)/4 which is around 5.3327.

Feb 9, 2024
#2
+1624
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Nice! Heron's Formula is a very important formula: The area of a triangle with side lengths $$a$$, $$b$$, and $$c$$, with semi-perimeter = $$a + b + c\over{2}$$ is:

$$\sqrt{s(s-a)(s-b)(s-c)}$$

This formula will come in handy, especially if you are doing competition math.

proyaop  Feb 9, 2024