In the adjacent figure, ∠ B = 90 degrees and D and E are the trisection points of BC then prove that 3AC^2 + 5AD^2 = 8AE^2 ?

In the adjacent figure,$\angle$ B = 90 degrees and D and E are the trisection points of BC then prove that $3 A {C}^{2}$ + $5 A {D}^{2}$ = $8 A {E}^{2}$

Jan 12, 2016

Use the Pythagorean Theorem and combine the equations

Explanation:

Calling $B C = 3 x$ => $B D = D E = C E = x$

${\triangle}_{A B D} \to A {D}^{2} = {x}^{2} + A {B}^{2}$ [1]
${\triangle}_{A B E} \to A {E}^{2} = 4 {x}^{2} + A {B}^{2}$ [2]
${\triangle}_{A B C} \to A {C}^{2} = 9 {x}^{2} + A {B}^{2}$ [3]

Subtracting [1] from [2]
$A {E}^{2} - A {D}^{2} = 3 {x}^{2}$ [4]

Subtracting [1] from [3]
$A {C}^{2} - A {D}^{2} = 8 {x}^{2}$ => ${x}^{2} = \frac{A {C}^{2} - A {D}^{2}}{8}$ [5]

Using [5] in [4]
$A {E}^{2} - A {D}^{2} = \left(\frac{3}{8}\right) \left(A {C}^{2} - A {D}^{2}\right)$
$8 A {E}^{2} - 8 A {D}^{2} = 3 A {C}^{2} - 3 A {D}^{2}$
$8 A {E}^{2} = 3 A {C}^{2} + 5 A {D}^{2}$
Q.E.D.