Question

In a triangle AD is perpendicular to BC and BD=1/3CD. Prove that 2AC2=2AB2+BC2

Answer 1

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Answer 2

In triangle ABC, let AD be perpendicular to BC and let BD = 1/3 CD.

Using the Pythagorean theorem on the right triangle ABD, we have:

$AB^2 + BD^2 = AD^2$

Since BD = 1/3 CD, we have:

$AB^2 + (1/3 CD)^2 = AD^2$

Expanding the square of 1/3 CD, we get:

$AB^2 + 1/9 CD^2 = AD^2$

Using the Pythagorean theorem on right triangle ACD, we have:

$AC^2 + AD^2 = CD^2$

Substituting the value of AD^2 from the equation derived above, we get:

$AC^2 + (AB^2 + 1/9 CD^2) = CD^2$

Expanding $AC^2$, we get:

$AC^2 = CD^2 - (AB^2 + 1/9 CD^2)$

Rearranging and substituting CD^2 = BC^2, we get:

$2AC^2 = 2BC^2 + 2AB^2 - 2/9 BC^2$

Cancelling out 2/9 $BC^2$ from both sides, we get:

$2AC^2 = 2AB^2 + BC^2$

This proves that $2AC^2 = 2AB^2 + BC^2.$

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