ODT2: All triangles are isosceles

== To Prove == All Triangles are isosceles Consider an arbitrary triangle ΔABC, shown in the figure below. Our goal is to show that AB = AC. This should be sufficient to establish that all triangles are isosceles. == Construction == Draw angle bisector of ∠A (AO) and Perpendicular bisector of side BC (DO). Let them intersect at O. Join point O to vertex of the triangle B and C (OB and OC respectively). Drop perpendicular from O on sides AB and AC (OE and OF respectively) The resulting diagram should look as follows. == Claim 1 == ΔAOE ≜ ΔAOF Proof: ∠OEA = ∠OFA ( = 900­­ ) ∠EAO = ∠FAO (AO is angle bisector of ∠A) AO = AO (Common) By AAS congruency, ΔAOE ≜ ΔAOF == Claim 2 == ΔBOD ≜ ΔCOD Proof:BD = CD (OD is perpendicular bisector of AD) ∠BDO = ∠CDO ( =900 ) DO = DO (Common) By SAS congruency, ΔBOD ≜ ΔCOD == Claim 3 == ΔEOB ≜ ΔFOC Proof: ∠OEB = ∠OFC( =900 ) BO = CO (ΔBOD ≜ ΔCOD as per claim 2) EO = FO (ΔAOE ≜ ΔAOF as per claim 1) By RHS congruency, ΔEOB ≜ ΔFOC == Claim 4 == AB = AC Proof: AE = AF(ΔAOE ≜ ΔAOF as per claim 1) EB = FC (ΔEOB ≜ ΔFOC as per claim 3) AE + EB = AF + FC(Adding last two lines) AB = AC Claim 4 establishes most of the proof. There are few more scenarios we need to consider before we can call the proof to be complete. <<<< Alternative Scenario 1 >>>> In construction step 1, we assumed that angle bisector of ∠A and perpendicular bisector of side BC intersect each other. Consider the situation where angle bisector of side A is perpendicular to side BC and intersects BC and D’. On other hand, perpendicular bisector of side BC is DA’. Consider ΔAD’B and ΔAD’C∠BAD’ = ∠CAD’(AD’ is the angle bisector) AD’ = AD’ (Common) ∠AD’B = ∠AD’C ( =900 ) From ASA congruency, ΔAD’B ≜ ΔAD’C This means, AD’ = D’C. This implies D’ is midpoint of BC. However midpoint of BC was assumed to be D (as A’D was perpendicular bisector). This means D and D’ are identical. This also means AD’ and AD are identical lines. Most important of all, AB = AC. This triangle is isosceles. <<<< Alternative Scenario 2 >>>> What if the angle bisector and perpendicular bisector are almost parallel? In this case point O will lie below the line BC. I could have asked reader to complete this part of the proof, however I will do it myself for sake of completeness. == Construction == Draw angle bisector of ∠A (AO) and Perpendicular bisector of side BC (DO). Let them intersect at O. Join point O to vertex of the triangle B and C (OB and OC respectively). Drop perpendicular from O on sides AB and AC (OE and OF respectively) In this case the perpendicular dropped from O on line AB and BC will lie on extended part of the respective line. Apart from that, all the constructions are identical to previous cases. == Claim 1 == ΔAOE ≜ ΔAOF Proof: ∠OEA = ∠OFA ( = 900­­ ) ∠EAO = ∠FAO (AO is angle bisector of ∠A) AO = AO (Common) By AAS congruency, ΔAOE ≜ ΔAOF == Claim 2 == ΔBOD ≜ ΔCOD Proof: BD = CD(OD is perpendicular bisector of AD) ∠BDO = ∠CDO ( =900 ) DO = DO (Common) By SAS congruency, ΔBOD ≜ ΔCOD == Claim 3 == ΔEOB ≜ ΔFOC Proof: ∠OEB = ∠OFC ( =900 ) BO = CO(ΔBOD ≜ ΔCOD as per claim 2) EO = FO (ΔAOE ≜ ΔAOF as per claim 1) By RHS congruency, ΔEOB ≜ ΔFOC == Claim 4 == AB = AC Proof: AE = AF(ΔAOE ≜ ΔAOF as per claim 1) EB = FC(ΔEOB ≜ ΔFOC as per claim 3) AE - EB = AF - FC (From last two lines) AB = AC This completes the proof. == Corollary == All triangles are equilateral. We have proved AB = AC. Using similar procedures we can show that AB = BC. This would mean AB = BC = CA. Thus all triangles are equilateral. Over to my esteemed opponent. (My apologies for poor formatting. I was having several problems with DDO text editor while copyinh it from word. )