In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AC - BC = 4$ and $BC - AB = 4$,find the lengths of all the sides of $\Delta ABC$.
(A) Let $AB = x$.
Given $BC - AB = 4$,so $BC = x + 4$.
Given $AC - BC = 4$,so $AC = BC + 4 = (x + 4) + 4 = x + 8$.
Since $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$,by the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
Substituting the values: $x^2 + (x + 4)^2 = (x + 8)^2$.
$x^2 + x^2 + 8x + 16 = x^2 + 16x + 64$.
$x^2 - 8x - 48 = 0$.
$(x - 12)(x + 4) = 0$.
Since $x$ must be positive,$x = 12$.
Thus,$AB = 12$,$BC = 12 + 4 = 16$,and $AC = 16 + 4 = 20$.
Given $BC - AB = 4$,so $BC = x + 4$.
Given $AC - BC = 4$,so $AC = BC + 4 = (x + 4) + 4 = x + 8$.
Since $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$,by the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
Substituting the values: $x^2 + (x + 4)^2 = (x + 8)^2$.
$x^2 + x^2 + 8x + 16 = x^2 + 16x + 64$.
$x^2 - 8x - 48 = 0$.
$(x - 12)(x + 4) = 0$.
Since $x$ must be positive,$x = 12$.
Thus,$AB = 12$,$BC = 12 + 4 = 16$,and $AC = 16 + 4 = 20$.
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