Dividing a triangle

Geometry Level 2

As shown, a triangle is divided into 7 parts--4 triangles and 3 quadrilaterals--by three line segments each drawn from a vertex to a point on the opposite side.

If the 4 triangles all have the same area, is it true that the areas of the three quadrilaterals are all equal?

Yes, it is always true No, it is always false No, it is only sometimes true

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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2 solutions

We will show that [ F G B ] = [ H G B ] [FGB]=[HGB] . Since they have a common altitude. Similarly △ H G B , △ H G I \triangle HGB, \triangle HGI have a common altitude, so: [ F G B ] [ F G A ] = F B F A , [ H G B ] [ H G I ] = H B H I \dfrac{[FGB]}{[FGA]}=\dfrac{FB}{FA}, \dfrac{[HGB]}{[HGI]}=\dfrac{HB}{HI} Similarly [ F I B ] [ F I A ] = F B F A , [ H A B ] [ H A I ] = H B H I \dfrac{[FIB]}{[FIA]}=\dfrac{FB}{FA}, \dfrac{[HAB]}{[HAI]}=\dfrac{HB}{HI} The △ F I B , △ H A B \triangle FIB,\triangle HAB have the same area, because both of them is bigger with a small triangle, than the F G H B FGHB quadrilateral. The △ F I A , △ H A I \triangle FIA,\triangle HAI have the same area too, because both of them is bigger with a small triangle, than △ A G I \triangle AGI . From that: F B F A = H B H I \dfrac{FB}{FA}=\dfrac{HB}{HI} And also ture, that [ F G B ] [ F G A ] = [ H G B ] [ H G I ] \dfrac{[FGB]}{[FGA]}=\dfrac{[HGB]}{[HGI]} Since in the fractions the denominators are equal, the counters are equal and [ F G B ] = [ H G B ] [FGB]=[HGB] The B G BG segment divides the △ A B D \triangle ABD to two equal parts, because [ F G A ] = [ H D B ] [FGA]=[HDB] and [ F G B ] = [ H G B ] [FGB]=[HGB] . So B G BG is a median, A G = G D AG=GD . From that C G CG is median too. Since [ G H I ] = [ E I C ] , [ D C I H ] = [ A G I E ] [GHI]=[EIC], [DCIH]=[AGIE] .

The proof was obtained with a proper letter change that all three quadrilaterals areas are equal.

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Marta Reece
Jul 4, 2017

It is obviously true for an equilateral triangle.

Any triangle can be obtained from an equilateral triangle by a linear transformation.

If two areas are equal to each other, linear transformation preserves that relationship.

The areas of the quadrilaterals are equal to each other in any triangle.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

i would put "and vice versa" on your second statement, i know it's even more trivial with the initial sentence but it does make the sol more complete .

Relue Tamref - 3 years, 10 months ago

Hmm, why is it obviously true for an equilateral triangle?

In that scenario, is it true that the 4 triangles have equal area if and only if the sides are cut at the same length? (If is obvious to me, but not only if.)

Calvin Lin Staff - 3 years, 10 months ago

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