+18 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!
+743 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 angle ADB is right, then so is angle ADC
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) =
(4) ( 4) (24/7) =
384/7 units^2
See the figure below to get a feel for this :
