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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 $$ \frac{a^2}{b^2} = \frac{x}{y} $$

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Here's a MathJax tutorial :) – Shaun Jul 9 '14 at 9:55
Dear Sir, Please give the complete solution – Achari S Ganesha Jul 9 '14 at 15:53
@user86508. It's not this site's policy to provide complete solutions to simple problems. We expect you to do some of the work yourself. deepu's solution is fine; if there's a step you don't understand, ask for clarification in a comment. – Rick Decker Jul 9 '14 at 17:58

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