Comments by "No Thanks" (@nothanks9503) on "Jake Broe"
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@JOE_XD Let us suppose, there’s a triangle with AB = 4, AC = 5 and BC = 3.
Construct the angle bisector of ∠A and the perpendicular bisector of segment B.C.
Now, in the constructed figure:
AB = 4
AC = 5
So, the angle bisector and perpendicular bisector are not parallel. Hence, they intersect at a point O. Drop perpendiculars OR and OQ to sides A.B. and A.C., respectively. Form segments O.B. and O.C.
Case 1:
AO = AO by reflexivity,
∠RAO = ∠QAO (AO is an angle bisector)
∠ARO = ∠AQO (both are right angles)
By A.A.S. congruence, ΔARO ≅ ΔAQO.
Consequently by CPCTC, AR = AQ and RO = OQ. ——-(1)
Case 2:
OD = OD by reflexivity,
∠ODB = ∠ODC (both are right angles)
BD = DC (OD bisects BC)
By S.A.S. congruence, ΔODB ≅ ΔODC.
Therefore, by CPCTC, O.B. = O.C. ——-(2)
Since we have proved that
R.O. = OQ ———-(1)
OB = OC ———-(2)
Also, since ∠O.R.B. and ∠O.Q.C. are both right angles, the hypotenuse-leg theorem for congruence implies ΔORB ≅ ΔOQC. Therefore, by CPCTC, B.R. = Q.C. —————(3)
We have shown that AR = AQ and BR = QC. Therefore, AB = AR + RB = AQ + QC = AC.
In other words, 4 = 5,
Thus, 2 + 2 = 5.
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