7/2/2018 Goal Beginning Proof • Proof Summary • We start our proof by drawing an arbitrary triangle • Prove that the centroid exists • Let our triangle be called ∆퐴퐵퐶 • Prove that the circumcenter exists • Let the midpoints of the triangle be 퐸, 퐹, 퐷 • Prove that the orthocenter lies on the line formed by the • This is important because: centroid and circumcenter • We want to show that each of the centers and the Euler • This completes the proof of the Euler Line Line exists for all possible triangles 28 29 Centroid Centroid • There is only one unique point on a median that • First, we arbitrarily draw the medians 퐶퐸 and 퐴퐹 splits the median into a 1:2 ratio (Midpoint-Vertex) • We could start with any 2 of the 3 medians • To show the centroid exists, we will show that: • Let 푀 be the intersection of 퐶퐸 and 퐴퐹 • Any two medians will always intersect at a point that splits the other median in a 1:2 ratio (Midpoint-Vertex) • Next, we draw the line 퐹퐸 • Hence, if the above statement is true, then the • We will show that 퐶푀 = 2푀퐸 and 퐴푀 = 2푀퐹 third median must intersect the other medians at the exact same point 30 31 Centroid Centroid • Consider ∆퐴퐵퐶 and ∆퐸퐵퐹 • Consider ∆퐴푀퐶 and ∆퐹푀퐸 • Since F is a midpoint, 퐵퐶 = 2퐵퐹 • By opposite angles, ∠퐸푀퐹 =∠퐴푀퐶 • Since E is a midpoint, 퐵퐴 = 2퐵퐸 • By alternate angles, ∠푀퐴퐶 =∠푀퐹퐸 • Also, ∠퐴퐵퐶 = ∠퐸퐵퐹 • By the AA rule, ∆퐴푀퐶~∆퐹푀퐸 • By the SAS rule, ∆퐴퐵퐶~∆퐸퐵퐹 • Since 퐴퐶 = 2퐹퐸 and ∆퐴푀퐶~∆퐹푀퐸, then 퐶푀 = • Since ∆퐴퐵퐶~∆퐸퐵퐹 and 퐵퐶 = 2퐵퐹, 퐴퐶 = 2퐹퐸 2푀퐸 and 퐴푀 = 2푀퐹 32 33 1 7/2/2018 Centroid Circumcenter • Thus, we have shown that: • Any two medians will always intersect at a point that • Perpendicular bisector of a line segment is the set splits the other median in a 1:2 ratio (Midpoint-Vertex) of all points which are equidistant from the • Hence, the third median 퐵퐷 must pass through 푀 endpoints of that line segment • Notice how point 푀 splits all three medians in a 1:2 ratio • If point 푌 lies somewhere on the perpendicular bisector (Midpoint-Vertex) of 퐵퐶, then 퐵푌 = 퐶푌 • Thus, the centroid exists and indeed it is point 푀 • If point 푍 is a point such that 퐵푍 = 퐶푍, then 푍 must lie on the perpendicular bisector of 퐵퐶 34 35 Circumcenter Circumcenter • First, we arbitrarily draw two perpendicular • Since 푅 lies on ℒ, 퐴푅 = 퐵푅 bisectors starting from midpoints 퐸 and 퐹 • Since 푅 lies on ℒ, 퐵푅 = 퐶푅 • Let ℒ be the perpendicular bisector passing through 퐸 • Since, 퐴푅 = 퐵푅 and 퐵푅 = 퐶푅, then 퐴푅 = 퐶푅 • Let ℒ be the perpendicular bisector passing through 퐹 • Let 푅 be the intersection of ℒ and ℒ 36 37 Circumcenter Orthocenter • Since 퐴푅 = 퐶푅, then 푅 must lie on the • We will use the centroid and circumcenter that we perpendicular bisector of the line segment 퐴퐶 proved previously to show that the orthocenter • Hence, the third perpendicular bisector must also lies on the same line. always intersect at the same point that the other • Recall: The centroid M splits the medians in a 2:1 ratio two perpendicular bisectors intersect. • Recall: 푅퐹, 푅퐷, 푅퐸 are the perpendicular bisectors • Thus, the circumcenter exists and indeed it is 푅 38 39 2 7/2/2018 Orthocenter Orthocenter • Consider ∆퐹푀푅 and ∆퐴푀푂 • By construction, 푂푀 = 2푀푅 • 푅푀 푂 푀푂 = 2푅푀 Extend line to a point such that • From the centroid proof, 퐴푀 = 2푀퐹 • We want to use the property of the centroid later on • By opposite angles, ∠퐹푀푅 = ∠퐴푀푂 • Draw the line segments 퐴푂, 퐵푂, 퐶푂 • Hence, by SAS rule, ∆퐹푀푅~∆퐴푀푂 • We will show that 푂 is indeed the orthocenter 40 41 Orthocenter Orthocenter • Consider ∆푅푀퐸 and ∆푂푀퐶 • Since ∆퐹푀푅~∆퐴푀푂, ∠푀퐹푅 =∠푀퐴푂 • By construction, 푂푀 = 2푀푅 • Since ∠푀퐹푅 = ∠푀퐴푂, 퐴푂 ∥ 푅퐹 (alternate angles) • From the centroid proof, 퐶푀 = 2푀퐸 • Since 푅퐹 ⊥ 퐵퐶 and 퐴푂 ∥ 푅퐹, then 퐴푂 ⊥ 퐵퐶 • By opposite angles, ∠푅푀퐸 = ∠푂푀퐶 • Hence, 퐴푂 is an altitude of the triangle • Hence, by SAS rule, ∆푅푀퐸~∆푂푀퐶 42 43 Orthocenter Orthocenter • Consider ∆푅푀퐷 and ∆푂푀퐵 • Since ∆푅푀퐸~∆푂푀퐶, ∠푀퐸푅 =∠푀퐶푂 • By construction, 푂푀 = 2푀푅 • Since ∠푀퐸푅 = ∠푀퐶푂, 퐶푂 ∥ 푅퐸 (alternate angles) • From the centroid proof, 퐵푀 = 2푀퐷 • Since 푅퐸 ⊥ 퐴퐵 and 퐶푂 ∥ 푅퐸, then 퐶푂 ⊥ 퐴퐵 • By opposite angles, ∠푅푀퐷 = ∠푂푀퐵 • Hence, 퐶푂 is an altitude of the triangle • Hence, by SAS rule, ∆푅푀퐷~∆푂푀퐵 44 45 3 7/2/2018 Orthocenter Orthocenter • Since ∆푅푀퐷~∆푂푀퐵, ∠푀푅퐷 =∠푀푂퐵 • Since ∆푅푀퐷~∆푂푀퐵, ∠푀푅퐷 =∠푀푂퐵 • Since ∠푀푅퐷 = ∠푀푂퐵, 퐵푂 ∥ 푅퐷 (alternate angles) • Since ∠푀푅퐷 = ∠푀푂퐵, 퐵푂 ∥ 푅퐷 (alternate angles) • Since 푅퐷 ⊥ 퐴퐶 and 퐵푂 ∥ 푅퐷, then 퐵푂 ⊥ 퐴퐶 • Since 푅퐷 ⊥ 퐴퐶 and 퐵푂 ∥ 푅퐷, then 퐵푂 ⊥ 퐴퐶 • Hence, 퐵푂 is an altitude of the triangle • Hence, 퐵푂 is an altitude of the triangle 46 47 Orthocenter Euler Line • But also, the orthocenter lies along the line 푅푀 • Since 퐴푂, 퐵푂, 퐶푂 are all altitudes of the triangle which we constructed earlier and 푂 is the point where all 3 altitudes intersect, then 푂 must be the orthocenter • Hence, the Euler Line exists and it is indeed 푅푀푂 • In addition, we know from how we constructed this line • Hence, the orthocenter exists and is indeed point 푂 that 푀푂 = 2푅푀 48 49 Recap So what? • Proof Summary • What are the applications of the Euler Line? • Prove that the centroid exists • Well, you now know something most of your peers don’t • We did this using the ratio 2:1 • University-Level Mathematics • Prove that the circumcenter exists • Applying mathematical concepts to real-life problems • We did this using the definition of perpendicular bisectors • Understanding and writing proofs about theorems • Prove that the orthocenter lies on the line formed by the centroid and circumcenter • Keys to Success • We did this by extending the circumcenter-centroid line • Try to generalize concepts segment to twice its original length • Try to explain and teach what you learned • This completes the proof of the Euler Line • Know when to study and when to take a break 50 51 4.