The sum of the hypotenuse and the altitude of a right angled triangle dropped on the hypotenuse exceeds the half-perimeter of the triangle.
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Let p be the length of perpendicular dropped from the right angle on the hypotenuse of length c of a right angled triangle of sides a and b. Then pc = ab = (twice the area of Δ) Also a2+b2=c2……(2) Now we have to prove p+c>a+b+c2 or 2p>a+b−c or 2.abc>a+b−c, by (1) or 2ab>ac+bc−c2=ac+bc−a2−b2 by (2) or a2+b2+2ab>c(a+b) or (a+b)2>c(a+b) or a+b>c[∵a+b>0] which is true since the sum of two sides of a triangle is greater than the third side.