SOLUTION: ABC is a triangle right angled at B let D and E be any points on AB and BC respectively prove that AE˛+CD˛ = AC˛+DE˛

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Question 1034944: ABC is a triangle right angled at B let D and E be
any points on AB and BC respectively prove that
AE˛+CD˛ = AC˛+DE˛
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

ΔABC is a right triangle, so by the Pythagorean
theorem, 

(1)     AB˛+BC˛ = AC˛


ΔABE is a right triangle, so by the Pythagorean
theorem, 

(2)     AB˛+BE˛ = AE˛.

ΔDBC is a right triangle, so by the Pythagorean
theorem, 

(3)     DB˛+BC˛ = CD˛.

ΔDBE is a right triangle, so by the Pythagorean
theorem, 

(4)     DB˛+BE˛ = DE˛.

Add BE˛+DB˛  to both sides of equation (1)


        AB˛+BC˛+BE˛+DB˛ = AC˛+BE˛+DB˛

Rearrange the terms:

        AB˛+BE˛+DB˛+BC˛ = AC˛+DB˛+BE˛

To make things easier to see, let's put parentheses 
around the first two terms on the left, the last two 
terms on the left, and the last two terms on the right:

(5)   (AB˛+BE˛)+(DB˛+BC˛) = AC˛+(DB˛+BE˛)
 
Using (2) we replace (AB˛+BE˛) in (5) by AE˛ 

Using (3) we replace (DB˛+BC˛) in (5) by CD˛

Using (4) we replace (DB˛+BE˛) in (5) by DE˛

And we end up with what we were to prove:

       AE˛+CD˛ = AC˛+DE˛
       
Edwin
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