1/7 triangle

The 'famous' 1/7 triangle.
We get it by joining the vertices of any ABC triangle to the point at 1/3 of opposite side.
The area of central triangle IJK is then 1/7 of that of ABC.
There are several analytic ways to prove it, however quite tedious.
We deal here with "visual proofs" and there correctness.

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A first "proof" is to draw the parallels from I,J,K and A,B,C to sides of IJK.
We get then 6 green triangles all equal to triangle IJK.
By symmetry, ADW is equal to BKN, and MEN equal to MJW.
The green "bits" outside ABC can then be put inside, and the overall area of the 7 green and yellow triangles is equal to ABC, hence Area(IJK) = 1/7 Area(ABC).

... but we need to prove these triangles are equal, that is N is the "other 1/3" of AB.
And here fails the "visual proof" as it is as hard than prooving directly by analytic ways that Area(IJK) = 1/7 Area(ABC) !

Really this property (M midpoint of AB and N at 1/3 of AB) is true only because the initial ratio is 1/3, only solution of 2r = 1 - r.
The value 1/3 should then be an explicit part of the proof that this visual construction is correct.
The previous applet shows this property : by moving point U at a ratio ≠ 1/3, we can see the disturbance on that "visual proof".
Point T is the key of an elementary proof, by using repeatedly the Thales theorem. The aim being to prove AJ = AK/2.

A quite "direct" proof (without tedious calculations) :
U is the transformed of C in the dilation (homothecy) centered in B with ratio 1/3, that we'll write U = HB, 1/3(C).
Similarily C = HV, -1/2(A).
The composition of these two dilations is a dilation with ratio (1/3)(-1/2) = -1/6
(It can't be a translation as ≠ 1) The center of this dilation is in line with the centers B and V.
This dilation changing A into U, its center is also in line with A,U.
The center of this dilation is then K. That is U = HK, -1/6(A).

Similarily, composition of dilations from centers C and W results into :
U = HC, 2/3(B) and B = HW, -2(A), hence U = HJ, -4/3(A).
That is, written without the signs as KU/1 = KA/6 = AU/7 and JU/4 = JA/3 = AU/7, obviously shows that J is the midpoint of AK (AJ = 3AU/7 and AK = 6AU/7).
The parallel projection on side AB shows then that M is midpoint of AB, and W midpoint of AN etc...

Other visual proof

The visual proof given at Cut-The-Knot is "just count the elementary triangles"... but these triangles are equal only for ratio = 1/3 !
Hence this (triangles are equal and other "visual evidence" of concurrent lines) should be carefully proved, as above (that KU = AU/7).

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Another variant

Still on the same relations, a proof by disection.

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Points D,E,F are here constructed as the symmetric from K,I,J to the midpoints M,N,P of the sides.
Here also, the "proof" is valid only after carefully prooving that the concerned polygons are parallelograms and the triangles are equal.
The figure is completely distorted and doesn't prove anything if the ratio is not 1/3 (move U).


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