If the sides of a given Triangle, taken cyclically, be divided in extreme and mean ratio; and if the points be joined: find the ratio which the area of the Triangle, so formed, has to the area of a given Triangle.
7 − 3√5.
Let ABC be the given Triangle; and let its sides be divided internally at A′, B′, C′, in extreme and mean ratio.
And let M be the area of ABC.
Let BA′ = x; then x² = a · (a − x);
i.e. x² + ax − a² = 0;
∴ x =
(−a ± a√5)/ |
(2) |
(a)/ |
(2) |
Then area of Triangle AB′C′
(c)/ |
(2) |
(b)/ |
(2) |
(1)/ |
(8) |
Similarly for BC′A′ and CA′B′.
Hence the sume of these 3 Triangles = 3 · (√5 − 2) · M, and area of Triangle A′BC′ = (7 − 3√5) · M.
Q.E.F.