A New Theorem introduced by Piyush Goel with Four Proof Mathematics for Piyush is a Passion from his childhood he was so passionate about Mathematics used to play with Numbers draw figures and try to get sides distance one day I draw a AP SERIES Right Angle Triangle( thinking that the distance between the point of intersection of median & altitude at the base must be sum of rest sides that was in My Mind) . And at last Piyush Succeed. This new Theorem proved with Four Proof(Trigonometry/Co-ordinates Geometry/Acute Theorem/Obtuse Theorem). Here are the Proofs: This Theorem applies in Two Conditions: 1. The Triangle must be Right-Angled. 2. Its Sides are in A.P. Series. 1. Proof with Trigonometry 2. Proof with Obtuse Triangle Theorem 3. Proof with Acute Triangle Theorem 4. Proof with Co-ordinates Geometry Four Proof ( TRIGONOMETRY/CO-ORDINATES/OBTUSE TRIANGLE/ACUTE TRIANGLE) (By PIYUSH GOEL) Theorem: In a Right-Angled Triangle with sides in A.P Theorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median & altitude at the base is 1/10 Th the sum of other two sides. This Theorem applies in Two Conditions: 1. The Triangle must be Right-Angled. 2. Its Sides are in A.P. Series. 1. Proof with Trigonometry tan α =AD/DC AD= DC tan α -----------------1 tan α = AD/DE AD= DE tan 2 α ----------------2 DC tan α = DE tan 2 α (DE+EC) tan α = DE tan 2 α DE tan α + EC tan α = DE tan 2 α DE tan α + EC tan α = 2 DE tan α / (1- tan^2 α ) DE tan α - DE tan^3 α + EC tan α – EC tan^3 α = 2 DE tan α EC tan α – EC tan^3 α - DE tan^3 α = 2 DE tan α – DE tan α tan α (EC – EC tan^2 α – DE t an^2 α )= DE tan α DE tan^2 α - DE = EC tan^2 α – EC -DE ( tan^2 α + 1) = -EC (1 - tan^2 α ) DE ( sin^2 α / cos^2 α + 1) = EC (1- sin^2 α / cos^2 α ) DE ( sin^2 α + cos^2 α / cos^2 α ) = EC ( cos^2 α - sin^2 α / cos^2 α ) DE ( sin^2 α + cos^2 α ) = EC( cos^2 α - sin^2 α ) DE ( sin^2 α + cos^2 α ) = EC ( cos^2 α - sin^2 α) ........where ( sin^2 α + cos^2 α =1) & ( cos^2 α - sin^2 α = cos 2 α ) DE= EC cos 2 α cos α =a/a+d & sin α = (a-d)/ (a +d) cos^2 α = a 2 / (a +b) 2 sin^2 α = (a-d) 2 / (a+ d) 2 DE= EC ( cos^2 α – sin^2 α ) = EC ( a 2 / (a +b) 2 - (a-d) 2 / (a +d) 2 = EC ( a 2 - (a-d) 2 / (a +d) 2 = EC (a – a +d) (a+ a-d)/ (a+ d) 2 = EC (d) (2 a -d)/ (a+ d) 2 = (a +d)/2(d) (2 a -d)/ (a +d) 2 ------------- where EC= (a +d)/2 = (d) (2 a -d)/2(a +d) = (d) (8 d -d)/2(4 d+d) ------------------where a= 4 d (as per the Theorem) = 7 d 2 /2 (5 d) = 7 d /10 = (3 d+4 d)/10= (A B+AC)/10 2. Proof with Obtuse Triangle Theorem AC 2 =EC 2 +A E 2 +2 C E. DE where EC = ( a +d) /2,A E=( a +d)/2 a 2 = (a +d/2) 2 + (a+ d/2) 2 + 2(a +d)/2 DE = (a +d/2) (a+d+2 DE) = (a +d/2) (a+d+2 DE) where a= 4 d 16 d 2 = (5 d/2) (5 d+2 DE) 32 d/5 = 5 d + 2 DE 32 d/5 - 5 d = 2 DE 32 d -25 d/5 = 2 DE DE =7 d /10 = (3 d+4 d)/10 = (A B+AC)/10 3. Proof with Acute Triangle Theorem A B 2 = AC 2 +BC 2 – 2 BC.DC (a-d) 2= a 2 + (a+ d) 2 -2(a+ d) (DE+EC) where A B= (a-d), AC=a, BC =( a +d) & EC= (a +d)/2 (a-d) 2 – (a +d) 2 = a 2 -2(a +d)(DE+EC) (a- d – a-d) (a -d +a +d) = a 2 -2(a+ d) (2 DE+a+d)/2 2(-2 d) (2 a) = 2 a 2 -2(a +d) (2 DE+a+d) -8 ad - 2 a 2 = -2(a +d) (2 DE+a+d) -2 a (4 d +a) = -2(a +d) (2 DE+a+d) a (4 d + a) = (a +d)(2 DE+a+d) 4 d (4 d + 4 d) = (4 d+d) (2 DE+4 d+d) 4 d (8 d ) = (5 d) (2 DE+5 d) 32 d 2 /5 d = (2 DE+5 d) 32 d/5 = (2 DE+5 d) 32 d/5 – 5 d = 2 DE (32 d - 25 d)/5 = 2 DE DE = 7 d/10 = (3 d+4 d)/10 = (AB+AC)/10 4. Proof with Co-ordinates Geometry In Triangle A B C,point A,B&C,s co-ordinates are respectively (0,0) ,(a,0)&(0,b). Point D is middle point ,co-ordinates of Point Dis (a/2,b/2) Equation of BE is .................. ( Two Points equation) Y – Y1 =(X – X 1 )(Y2 – Y1)/(X2 – X 1 ) Y – 0 =b-0/0-a(X – a) Y = -b/a(X) + b------------------- (1)
