Within a given Triangle place a similar Triangle, whose area shall have to its are a given ratio less than unity, whose sides shall be parallel to its sides, and whose vertices shall be equidistant from its vertices.
Let ABC be the given Triangle, and A′B′C′ the required one; and let the ratio, which B′C′ has to BC, be ‘k’; so that k is less than 1.
Since BB′ = CC′, and that BC, B′C′, are parallel, it may easily be proved, by dropping perpendiculars from B′, C′, upon BC, which must necessarily be equal, that ∠s B′BC, C′CB, are equal.
Similarly, ∠s A′AC, C′CA are equal; and so are ∠s A′AB, B′BA.
Call ∠B′BC ‘θ’; then ∠C′CB = θ
∴ ∠C′CA = C − θ = ∠ A′AC;
∴ ∠A′AB = A − (C − θ) = ∠B′BA.
Now ∠s B′BC, B′BA, together = B;
∴ θ + A − (C − θ) = B;
∴ 2θ = B + C − A = 180° − 2A;
∴ θ = 90° − A.
Hence, if BB′, CC′, be produced to meet at D, Triangle DBC will be isosceles, with a vertical ∠ equal to 2A.
Now, if a Circle be drawn about ABC, and its centre joined to B and C, the Triangle, so formed, will fulfil the same conditions;
hence the centre of this Circle will be D;
hence the construction.
Bisect the sides, and draw perpendiculars, meeting at D. Join D to the vertices B, C. From DB cut off DB′ = k · DB. From B′ draw B′C′ parallel to BC.
Then B′C′ is easily proved equal to k · BC.
And if, from B&prime, C′, parallels to AB, AC, be drawn, it may easily be proved that they meet on DA, and that they are respectively equal to k · AB, k · AC.