In a given acute-angled Triangle inscribe a Triangle, whose sides make, at each of the vertices, equal angles with the sides of the given Triangle.
Let ABC be the given Triangle, and A′B′C′ the required Triangle, so that ∠ BA′C′ = ∠CA′B′, &c.
Evidently, A′C′, A′B′ are equally inclined to a line drawn, from A′, ⊥; BC; and so of the others: i.e. these ⊥;s bisect the ∠s at A′, B′, C′;
∴ they meet in the same Point. Draw them; let them meet at O; and call the ∠C′A′B′ ‘2α’, and so on.
Now (β + γ) = π − ∠B′OC′ = A;
∴ 2A = 2(β + γ) = π − 2α
α = 90° − A;
∴ ∠BA′C′ = A.
Similarly ∠BC′A′ = C.
∴ Triangle BC′A′ is similar to Triangle BCA; and so on of the others;
∴ A′ is foot of ⊥ drawn, from A, to BC. Hence the construction is obvious.