# Pythagoras theorem and ratios of sides

In a right angle LAB angle L is right angle and AM is perpendicular to AB. Prove that LA^2 : LB^2 = AM : MB. This can be proved using similar triangle principles. But I am interested to prove this using only Pythagoras the

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Polish your question a bit. Please be a little more helpful by producing the diagram of the triangle in question. –  MonK Jul 9 '14 at 9:59
The problem as stated is incorrect. Presumably you meant that M lies on AB and that LM is perpendicular to AB. –  Rick Decker Jul 9 '14 at 17:55

Let $LA = a ,LB = b ; AM = x ,BM = y$
From Pythagoras' theorem on ALB, $$a^2 + b^2 = (x+y)^2 \tag{1}$$
From Pythagoras' theorem on AML and LMB , $$a^2-b^2 = x^2-y^2 \tag{2}$$ Adding equations (1) & (2) , we get $$a^2=x(x+y)$$ Subtracting equation (2) from (1) we have $$b^2=y(x+y)$$ From which we get $$\frac{a^2}{b^2} = \frac{x}{y}$$