In
,
and
.
Also the perpendicular bisector of side , the internal angle bisector of and the side are concurrent.
Find the area of .
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AB:AC = 2:3 and AP is the angle bisector hence AP:PC = 2:3. Take them to be 2x and 3x respectively.
PM perpendicular bisector of AB → P M = 4 x 2 − 1
Drop PQ perpendicular to AC.
Triangles APM and BPM are congruent giving BP = AP = 2x and AQ = 1; QC = 2
P C = 4 x 2 + 3 which was originally taken as 3x
Equate and solve for x getting x = 5 3
Area of ABC would be 2 ( 2 + 3 ) 4 5 3 − 1 = 2 3 5 = 2 . 9 5 8