Question 174528
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In triangle ABC ,angle ACB is 90 degrees and CD is perpendicular to AB.
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Let's draw the Triangle according to the condition:
{{{drawing(400,400,-6,1,-1,6,graph(400,400,-6,1,-1,6),triangle(0,0,-5,0,0,5),blue(line(-2.5,2.5,0,0)),blue(locate(-5.2,0.3,A)),blue(locate(0.2,5.2,B)),blue(locate(0.2,0.3,C)),green(line(-.5,0,-.5,.5)),green(line(-.5,.5,0,.5)),blue(locate(-2.7,2.8,D)),red(locate(-4.7,.5,45^o)),red(locate(-2.7,2.2,90^o)),red(locate(-1,.5,45^o)),red(locate(-.5,1,45^o)),red(locate(-.5,4.5,45^o)),red(locate(-2.2,2.8,90^o)))}}} -------> "{{{CD}}}" perpendicular to "{{{AB}}}"; Angle "{{{ACB=90^o}}}"
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In Trigonometric Function for Right Triangles, {{{tan(45^o)=1}}}, and we know {{{tangent=opp/adjacent}}}
And it shows in our Triangle, {{{tan(45^o)=highlight(AD)/CD=highlight(1)/1=1}}}
Also, {{{tan(45^o)=highlight(BD)/CD=highlight(1)/1=1}}}
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By Pyth. Theorem: ----> {{{hyp^2=opp^2+adj^2}}}
{{{CA^2=AD^2+CD^2=1^2+1^2=1+1}}}
{{{highlight(CA^2=2)}}}
Also,{{{CB^2=CD^2+BD^2=1^2+1^2=1+1}}}
{{{highlight(CB^2=2)}}}
Therefore,
Prove {{{CB^2/CA^2=BD/AD}}}, substitute highlighted values:
{{{2/2=1/1}}}
{{{1=1}}}
Thank you,
Jojo</pre>