## Knowledge in Mathematics Proof

Piyush Theorem

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;PiyushTheorem PiyushTheorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median &amp; altitude at the base is 1/10 Th&nbsp;the sum of other two sides. This Theorem applies in Two Conditions: 1.The&nbsp;Triangle&nbsp;must be&nbsp;Right-Angled. 2.Its&nbsp;Sides&nbsp;are in&nbsp;A.P. Series. 1.Proof with Trigonometry Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;=AD/DC AD= DC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&mdash;&mdash;&mdash;&mdash;&mdash;&ndash;1 Tan&nbsp;&nbsp;&alpha;&nbsp;= AD/DE AD= DE&nbsp;Tan2&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&mdash;&mdash;&mdash;&mdash;&mdash;-2 DC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= DE&nbsp;Tan 2&nbsp;&alpha; (DE+EC)&nbsp;&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= DE&nbsp;Tan 2&nbsp;&alpha; DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;+ EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= DE&nbsp;Tan 2&nbsp;&alpha; DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;+ EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;= 2 DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;/ (1-&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;) DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&ndash; DE&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;+ EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&ndash;EC&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;&nbsp;= 2DE&nbsp;Tan&nbsp;&nbsp;&alpha; EC&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&ndash;EC&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;&ndash; DE&nbsp;Tan3&nbsp;&nbsp;&alpha;&nbsp;= 2DE&nbsp;Tan&nbsp;&nbsp;&alpha;&nbsp;&ndash; DE&nbsp;Tan&nbsp;&nbsp;&alpha; Tan&nbsp;&nbsp;&alpha;&nbsp;(EC &ndash; EC&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&ndash; DE&nbsp;T an2&nbsp;&nbsp;&alpha;&nbsp;)= DE&nbsp;Tan&nbsp;&nbsp;&alpha; DE&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&ndash; DE = EC&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;&nbsp;&ndash; EC -DE (&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;+ 1) = -EC (1 &ndash;&nbsp;Tan2&nbsp;&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;&nbsp;/cos2&nbsp;&alpha;&nbsp;+ 1) = EC (1-&nbsp;sin2&nbsp;&alpha;&nbsp;&nbsp;/cos2&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;/cos2&nbsp;&alpha;&nbsp;) = EC (cos2&nbsp;&alpha;&nbsp;&ndash;&nbsp;sin2&nbsp;&alpha;&nbsp;/cos2&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;) = EC(cos2&nbsp;&alpha;&nbsp;&nbsp;&ndash;sin2&nbsp;&alpha;&nbsp;) DE (sin2&nbsp;&alpha;&nbsp;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;) = EC (cos2&nbsp;&alpha;&nbsp;&nbsp;&ndash;sin2&nbsp;&alpha;&nbsp;) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where (sin2&nbsp;&alpha;&nbsp;&nbsp;+&nbsp;cos2&nbsp;&alpha;&nbsp;=1) &amp; (cos2&nbsp;&alpha;&nbsp;&nbsp;&ndash;sin2&nbsp;&alpha;&nbsp;=&nbsp;cos2&nbsp;&alpha;&nbsp;&nbsp;) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;DE= EC&nbsp;cos2&nbsp;&alpha;&nbsp;&nbsp; cos&nbsp;&alpha;&nbsp;&nbsp;&nbsp;=a/a+d &nbsp;&nbsp;&amp;&nbsp;sin&nbsp;&alpha;&nbsp;= (a-d)/ (a +d) cos2&nbsp;&alpha;&nbsp;&nbsp;= a2/ (a +b)&nbsp;2 sin2&nbsp;&alpha;&nbsp;&nbsp;= (a-d)&nbsp;2/ (a+ d)&nbsp;2 DE= EC (cos2&nbsp;&alpha;&nbsp;&nbsp;&nbsp;&nbsp;&ndash;&nbsp;sin2&nbsp;&alpha;&nbsp;) = EC (a2&nbsp;/ (a +b)&nbsp;2&nbsp;&ndash; (a-d)&nbsp;2/ (a +d)&nbsp;2 = EC (a2&nbsp;&ndash; (a-d)&nbsp;2/ (a +d)&nbsp;2 = EC (a &ndash;a +d) (a+ a-d)/ (a+ d)&nbsp;2 = EC (d) (2a -d)/ (a+ d)&nbsp;2 = (a +d)/2(d) (2a -d)/ (a +d)&nbsp;2&nbsp;&mdash;&mdash;&mdash;&mdash;- where EC= (a +d)/2 = (d) (2a -d)/2(a +d) = (d) (8d -d)/2(4d+d) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;where a= 4d (as per the Theorem) = 7d2&nbsp;/2(5d) = 7d /10 = (3d+4d)/10= (AB+AC)/10 2.Proof with Obtuse Triangle Theorem AC2=EC2&nbsp;+AE2&nbsp;+2CE.DE&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where EC = ( &nbsp;a +d) /2,AE=( a +d)/2 a2&nbsp;= (a +d/2)2&nbsp;+ (a+ d/2)2&nbsp;+ 2(a +d)/2DE = (a +d/2) (a+d+2DE) = (a +d/2) (a+d+2DE) &nbsp;&nbsp;where a=4d 16d2&nbsp;= (5d/2) (5d+2DE) 32d/5 = 5d + 2DE 32d/5 &ndash; 5d = 2DE 32d -25d/5 = 2DE DE =7d/10 = (3d+4d)/10 = (AB+AC)/10 3.Proof with Acute Triangle Theorem AB2= AC2+BC2&nbsp;&ndash; 2BC.DC (a-d) 2= a2&nbsp;+ (a+ d)&nbsp;2&nbsp;-2(a+ d) (DE+EC) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;where AB= (a-d), AC=a, BC =( a +d) &amp; EC= (a +d)/2 (a-d)&nbsp;2&nbsp;&ndash; (a +d)2&nbsp;= a2&nbsp;&nbsp;-2(a +d)(DE+EC) (a- d &ndash;a-d) (a -d +a +d)&nbsp;&nbsp;= a2&nbsp;-2(a+ d) (2DE+a+d)/2 2(-2d) (2a)&nbsp;=&nbsp;2a2&nbsp;-2(a +d) (2DE+a+d) -8ad&nbsp;&ndash;&nbsp;2a2&nbsp;= -2(a +d) (2DE+a+d) -2a (4d&nbsp;&nbsp;&nbsp;+a) = -2(a +d) (2DE+a+d) a (4d&nbsp;&nbsp;+ a) = (a +d)(2DE+a+d) 4d (4 d&nbsp;+&nbsp;4d) = (4d+d) (2DE+4d+d) 4d (8d)&nbsp;= (5d) (2DE+5d) 32d2/5d = &nbsp;&nbsp;(2DE+5d) 32d/5 = &nbsp;&nbsp;(2DE+5d) 32d/5 &ndash; 5d = &nbsp;&nbsp;2DE (32d &ndash; 25d)/5 = &nbsp;&nbsp;2 DE DE = 7d/10 = (3d+4d)/10 = (AB+AC)/10 4. Proof with Co-ordinates Geometry Equation of BE Y &ndash; 0 =b-0/0-a(X &ndash; a) Y = -b/a(X) + b&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;- (1) M1&nbsp;= -b/a For perpendicular M1M2= -1 So M2=a/b Equation of AC Y &ndash; 0 = a/b(X-0) Y=a/b(X) &mdash;&mdash;&mdash;&mdash;&mdash;&mdash; (2) Put Y value in equation (1) a/b(X) + b/a(X) =b X (a2+b2/a b) = b X = ab2/ (a2&nbsp;+ b2) To get Value of Y, put X value in equation (2) Y = a/b (ab2/ (a2+b2) Y = a2b/ (a2+b2) Here we got co-ordinates of Point C &ndash; ab2/ (a2&nbsp;+ b2), a2b/ (a2+b2) and co-ordinates of point d is (a/2, b/2) because d is midpoint. As per the &ldquo;Theorem&rdquo; a=z-d, b=z, c = z+ d (z +d)&nbsp;2= (z-d)&nbsp;2+z2&nbsp;from here z=4d so a=3d and b=4d Put value of a &amp; b ab2/ (a2&nbsp;+ b2), a2b/ (a2+b2) &amp; (a/2, b/2) ab2/ (a2&nbsp;+ b2) = 48d/25 a2b/ (a2+b2) = 36d/25 a/ 2=3d/2 b/ 2 =4d/2 CD2= (48d/25 -3d/2)2-(36d/25-4d/2)2 = (96d-75d/50)2&nbsp;+ (72d-100d/50)2 = (21d/50)2&nbsp;+ (-28d/50)2 = (441d2/2500) + (784d2/2500) = (1225d2/2500) CD= 35d/50 = 7d/10 = 7d/10 = (3d+4d)/10 = (AB+AE)/10 https://piyushtheorem.wordpress.com/2017/02/08/a-theorem-on-right-angled-triangles/

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