Geometric Fallacy Proof: Every Acute Triangle Is Isosceles
FIND THE ERROR IN THE FOLLOWING PROOF:
Construction: In an acute triangle 4ABC construct an angle bisector
of the angle ∠C and the perpendicular bisector of the side AB. Denote the
point of intersection P . From the point P draw the perpendicular P U to the
side AC and the perpendicular P V to the side BC.
Thus, we have:
AM = M B;
∠ACP = ∠P CB;
P M ⊥ AB; P U ⊥ AC; P V ⊥ BC.
Proof:
(1). 4AM P = 4BM P by SAS (P M - common side, AM = M B and
∠AM P = ∠BM P = 90o .). Hence, AP = BP.
(2). 4CU P = 4CV P by AAS (CP - common side, ∠U CP = ∠P CV
by construction, and ∠CU P = ∠CV P = 90o .). Hence, U P = V P and
CU = CV.
(3). 4AU P = 4BV P by HL (U P = V P by (2), and AP = BP by
(1)). Hence, AU = BV.
(4). Since AU = BV (by (3)) and CU = CV (by (2)), we have that
AC = BC, and the triangle is isosceles.