ABC is a right-angled triangle. Construct a point P inside ABC so that the angles PAB, PBC, PCA are equal.
Solution
∠APB = 180o - ∠PAB - ∠PBA = 180o - ∠PBC - ∠PBA = 180o - ∠B = 90o. Similarly, ∠BPC = 180o - ∠C.
Taking P on the circle diameter AB ensures ∠APB = 90o. Take X as the intersection of the perpendicular to AC at C and the line AB. Then ∠BXC = ∠C, so if P lies on the minor arc BC, then ∠BPC = 180o - ∠C.
© John Scholes
jscholes@kalva.demon.co.uk
11 Oct 2003
Last corrected/updated 11 Oct 03