Geometric Fallacy Proof: ”Slightly obtuse” angle is equal to the right angle.
FIND THE ERROR IN THE FOLLOWING PROOF:
Construction: Construct the right angle ∠ADC and ”slightly obtuse”
angle ∠DAE (away from CD, with CD = AE. Construct the perpendicular
bisector of AD and the perpendicular bisector of CE. Since AD and CE are
not parallel, their perpendicular bisectors are also not parallel. Denote the
point of intersection of the perpendicular bisectors O.Connect point O with
the points A, D, C and E.
Thus, we have:
CD = AE; DH = AH; CK = KE;
OH ⊥ AD; OK ⊥ CE.
Proof:
(1). 4DHO = 4AHO by SAS (HO - common side, AH = HD and
∠DHO = ∠AHO = 90o .). Hence, DO = AO and ∠ODH = ∠OAH.
(2). 4COK = 4EOK by SAS (KO - common side, CK = EK and
∠CKO = ∠EKO = 90o .). Hence, CO = EO.
(3). 4CDO = 4EAO by SSS (CD = AE by construction; DO = AO
by (1), and CO = EO by (2)). Hence, ∠CDO = ∠EAO.
(4). Since ∠CDO = ∠EAO (by (3)) and ∠ODH = ∠OAH. (by (1)),
we have that ∠DAE = ∠ADC = 90o .