PROVE: <ABC = <BCA (i.e. triangle ABC is an isosceles triangle)
A
/ \
/ \
D E XP normal to AB
/ \ / \ XQ normal to AC
P /----X----\ Q
/ / \ \
/ / \ \
/ / \ \
B/_ _ _ ___ _\C
- PROOF
Let XP and XQ be normals to AC and AB. Since the three angle bisectors are concurrent, AX bisects angle A also and therefore XP = XQ.
Let's assume XD > XE. Then ang(PDX) < ang(QEX) Now considering triangles BXD and CXE,
the last condition requires that
ang(DBX) > ang(ECX)
OR ang(XBC) > ang(XCB) OR XC > XB
- Thus our assumption leads to
- XC + XD > XE + XB
OR CD > BE
which is a contradiction.
Similarly, one can show that XD < XE leads to a contradiction too.
Hence XD = XE => CX = BX From which it is easy to prove that the triangle is isosceles.
-- Manish S Prabhu (mprabhu@magnus.acs.ohio-state.edu)