Statement: In a Right-angled ◺ Triangle, if distance between middle point and perpendicular point and sum of other two sides is divisible by 10, then sides will be in A.P. Series.
As per Figure
Z^2=X^2 +Y^2
X^2 = AC^2 + BC^2 X^2 = AC^2 + (Z/2 – X+Y/10)
Y^2 = AC^2 + CE^2 Y^2 = AC^2 + (Z/2 + X+Y/10)
X^2 – Y^2 = AC^2 – AC^2 + (Z/2-X+Y/10 + Z/2 – X+Y/10)(Z/2 – X+Y/10 – Z/2 –X+Y/10)
(X+Y)(X – Y) = (Z) (- X+Y/5)
(X-Y) = (-Z/5)
Z = 5Y-5X …Put this value into
Z^2 = X^2 + Y^2
(5Y -5X)^2 = X^2 + Y^2
25Y^2 + 25 X^2 – 50XY = X^2 + y^2
24X^2 – 50XY + 24Y^2 = 0
24X^2 – 32XY – 18XY + 24Y^2 = 0
8X (3X – 4Y) – 6Y (3X – 4Y) =0
(3X -4y)(8X – 6Y) =0
2(3X -4Y) (4X – 3Y) =0
(3X -4Y)(4X – 3Y) =0
3X -4Y =0 3X =4Y 3/4 = Y/X or X= 4Y/3 or Y = 3X/4
4X – 3Y =0 4X = 3Y 4/3 = Y/X or X= 3Y/4 OR Y = 4X/3
Put X Value in this Equ 5Y -5X = Z
5*4X/3 – 5X =Z
20X – 15X = 3Z
5X =3Z
X/Z = 3/5
Y/X =4/3
So X =3, Y =4 & Z =5 that is A P Series.
QED