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# Theorem In a Right-Angled Triangle with sides in A.P

Piyush, born on 10th Feb, 1967, Aquarian belongs to a middle class family in Dadri, Near Noida, elder son of Dr. Devender Kumar Goel and mother  Ravikanta. He is Diploma Mechanical Engineering passed in the year 1987, Diploma in Material Management, Diploma in Vastu Shastra and Diploma in Business Management. creative, believe in God too much, believe in Love & Friendship.

Piyush is now known as ‘Mirror Image ManPiyush is very passionate about mathematics he has done lot work in Mathematics .
1.Amazing Number Nine
2. Squaring A New WAY
3. Piyush Formula And Table For Squaring
4.Squaring Through Squares
5.A New Method Of Squaring and Cubing
6.A New Theorem Introduced by Piyush Goel
7. How To Get 11 (Power 5) From Pascal Triangle
8.Counting 1,2,3 …..9 ,An Amazing
9.Factorial Function
Theorem: In a Right-Angled Triangle with sides in A.P by Piyush Goel
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:

1. ∆ABC is Right-Angled
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

a(a-4d) = 0

a – 4d = 0 (as a ≠ 0)

a = 4d (———-eqn. 1)

In ∆ABD

(a – d)2 = BD2 + AD2

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

In ∆ACD

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, AB = a –d = 4d – d = 3d & AC = a = 4d

Putting these values, we get

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

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