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# Help would be greatly appreciated

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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=5 then compute the area of triangle ABC.

Please explain your answer fully because I want to understand how to do it. Please do not use anything advanced like trigonometry as I don't understand trig and I wish to understand the problem. Thank you in advance!

Oct 12, 2023

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There is a nice soluiton here: https://web2.0calc.com/questions/hxlp

Since DI =3, BD = 4  and BI = 5, then triangle IDB  is right with angle IDB = ADB = 90°

And since  ID = 3  and BI = 5.....then sin (DBI)  = 3/5

And cos DBI = 4/5

But since BE is an angle bisector, then sin (ABC)  = sin (2 * DBI)  =  2 sin(DBI)cos(DBI) = 2 ( 3/5) (4/5) = 24/25

And ADB is also a right triangle with BD  = 4

And since sin ABC = 24/25  then  triangle ADB  is a 7 - 24- 25  Pythagorean right triangle

Then sin ABD = 24/25

So sin of DAB  = 7/25

Using the Law of Sines

AB/ sin 90 = BD / sin (DAB)

AB/ 1  = 4 / (7/25)  =  100/7

And

AD  / sin (ABC)  = BD / sin (DAB)

AD = sin (ABC) * BD / sin (DAB)

AD = (24/25) * 4  / ( 7/25)  = 96/7

And since AD is an  angle bisector then  sin (BAD)  = sin (DAC)  = 7/25

So sin (ACD)   = 24/25

So we can find DC using the Law of Sines

DC / sin (DAC)  = AD / sin (ACD)

DC = sin (DAC) * AD / sin (ACD)

DC = (7/25) * ( 96/7) / (24/25)

DC = (7/25) * (96/7) * (25/24)

DC = (96/24)  = 4

So  BC =  BD + DC  = 4 + 4  = 8

And  the area of triangle ABC =  (1/2) (BC) ( AB) * sin ( ABC)  = (1/2) ( 8) ( 100/7) ( 24/25)  =

384/7  units^2

See the figure below to  get a feel  for this : Oct 12, 2023