Question

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,`/_` B = 90 degrees and D and E are the trisection points of BC then prove that `3AC^2` + `5AD^2` = `8AE^2`

Answer 1

Use the Pythagorean Theorem and combine the equations

Explanation:

Calling `BC=3x` => `BD=DE=CE=x`

`triangle_(ABD) -> AD^2=x^2+AB^2` [1]
`triangle_(ABE) -> AE^2=4x^2+AB^2` [2]
`triangle_(ABC) -> AC^2=9x^2+AB^2` [3]

Subtracting [1] from [2]
`AE^2-AD^2=3x^2` [4]

Subtracting [1] from [3]
`AC^2-AD^2=8x^2` => `x^2=(AC^2-AD^2)/8` [5]

Using [5] in [4]
`AE^2-AD^2=(3/8)(AC^2-AD^2)`
`8AE^2-8AD^2=3AC^2-3AD^2`
`8AE^2=3AC^2+5AD^2`
Q.E.D.