A theorem on right angled triangles

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 1/10th 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.

We Have:

image

  1. ∆ABC is Right-Angled
  2. AD is Altitude
  3. AE is Median i.e. E is the midpoint of BC

Proof:

(a+d)2 = a2 + (a-d)2

(a+d)2 -(a-d)2 = a2

a2 + d2 + 2ad - a2 - d2 + 2ad = a2

4ad = a2

a(a-4d) = 0

a – 4d = 0 (as a ≠ 0)

a = 4d (----------eqn. 1)

 

In ∆ABD

AB2 = BD2 + AD2

(a - d)2 = BD2 + AD2

(a - d)2 = {(a + d)/2 – DE}2 + AD2 (----------eqn. 2)

In ∆ACD

AC2 = DC2 + AD2

a2 = DC2 + AD2

a2 = {(a + d)/2 + DE}2 + AD2 (----------eqn. 3)

 

From eqn. 2 & 3, we get

(a - d)2 - a2 = {(a + d)/2 – DE}2 + AD2 - {(a + d)/2 + DE}2 - AD2

(a - d +a )(a - d - a) = {(a+d)/2 – DE + (a+d)/2 + DE}{(a+d)/2 – DE - (a+d)/2 - DE}

(2a – d)(-d) = (a + d)(-2DE)

(2a – d)(d) = (a + d)(2DE)

So, 2DE = (2a - d)d/(a+d)

From eqn. 1, we get

2DE = (2*4d – d)d/(4d + d)

2DE = 7d2/5d

DE = 7d/10 = (4d + 3d)/10

But, AD = a –d = 4d – d = 3d & AC = a = 4d

Putting these values, we get

DE = (AC + AB)/10 (Hence Proved)

 

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Piyush Goel

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