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)

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