the medians of triangle abc intersects at G prove that ar(tri AGB)=ar(tri AGC)=ar(tri BGC)=1/3ar(tri ABC) - Maths - Areas of Parallelograms and Triangles

Question

the medians of triangle abc intersects at G prove that ar(tri
AGB)=ar(tri AGC)=ar(tri BGC)=1/3ar(tri ABC)

Ajanta Trivedi · Accepted Answer

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given: ABC is a triangle D , E and F are the midpoints of the
sides BC, AC and AB respectively medians intersect at G
TPT: area(Î”AGB) =
area(Î”AGC)=area(Î”BGC)=1/3
area(Î”ABC)
proof:
in the triangle ABC , AD is the median
area(Î”ABD) = area(Î”ACD) (1) [since
medians divide the triangle into two equal parts]
in triangle GBC, GD is the median
area(Î”GBD) = area(Î”GCD) (2)
subtract (2) from (1):
area(Î”ABD) - area(Î”GBD) =
area(Î”ACD) - area(Î”GCD)
area(Î”AGB) = area(Î”AGC) (3)
similarly we can show that area(Î”BGC) =
area(Î”AGC) (4)
from (3) and (4):
area(Î”AGB) = area(Î”AGC) =
area(Î”BGC)
area(Î”AGB) + area(Î”AGC) +
area(Î”BGC) = area(Î”ABC)
area(Î”AGB) = area(Î”AGC) =
area(Î”BGC) = 1/3 area(Î”ABC)
hope this helps you

the medians of triangle abc intersects at G. prove that ar(tri.AGB)=ar(tri.AGC)=ar(tri.BGC)=1/3ar(tri.ABC)