## 04 Nov 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 1/10**^{th}** the sum of other two sides****.**

**This Theorem applies in Two Conditions:**

- The
**Triangle**must be**Right-Angled**. - Its
**Sides**are in**A.P. Series**.

**We Have:**

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

**Proof:**

(a+d)^{2} = a^{2} + (a-d)^{2}

(a+d)^{2} -(a-d)^{2 }= a^{2}

a^{2} + d^{2} + 2ad – a^{2} – d^{2 } + 2ad = a^{2}

4ad = a^{2}

a(a-4d) = 0

a – 4d = 0 (as a ≠ 0)

a = 4d **(———-eqn. 1)**

In ∆ABD

AB^{2} = BD^{2} + AD^{2}

(a – d)^{2 }= BD^{2} + AD^{2}

(a – d)^{2 }= {(a + d)/2 – DE}^{2} + AD^{2 }**(———-eqn. 2)**

In ∆ACD

AC^{2} = DC^{2} + AD^{2}

a^{2} = DC^{2} + AD^{2}

a^{2} = {(a + d)/2 + DE}^{2} + AD^{2 }**(———-eqn. 3)**

From eqn. **2** & **3**, we get

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

(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 = 7d^{2}/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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