Question 1061271
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In any triangle ABC, E is any point on altitude line segment AD. Prove that (AC)^2-(CE)^2= (AB)^2-(EB)^2
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0.  Make a sketch to follow my arguments.


1.  From the triangle ADC:  {{{abs(DC)^2}}} = {{{abs(AC)^2 - abs(AD)^2}}}.

    From the triangle EDC:  {{{abs(DC)^2}}} = {{{abs(CE)^2 - abs(ED)^2}}}.

    It implies  {{{abs(AC)^2 - abs(AD)^2}}} = {{{abs(CE)^2 - abs(ED)^2}}}.

    Hence,      {{{abs(AC)^2 - abs(CE)^2}}} = {{{abs(AD)^2 - abs(ED)^2}}}.   (1)


2.  Similarly,
    From the triangle ADB:  {{{abs(BD)^2}}} = {{{abs(AB)^2 - abs(AD)^2}}}.

    From the triangle EDC:  {{{abs(BD)^2}}} = {{{abs(BE)^2 - abs(ED)^2}}}.

    It implies  {{{abs(AB)^2 - abs(AD)^2}}} = {{{abs(BE)^2 - abs(ED)^2}}}.

    Hence,      {{{abs(AB)^2 - abs(BE)^2}}} = {{{abs(AD)^2 - abs(ED)^2}}}.   (2)


3.  From (1) and (2) you have 

    {{{abs(AB)^2 - abs(BE)^2}}} = {{{abs(AC)^2 - abs(CE)^2}}}.


It is what has to be proved.
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Solved.