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Attachment:
6c4cOJzjJV-group-57.png
Triangle ABC is similar to triangle ADE

Hence, \(\frac{AC}{CE} = \frac{AB}{BD}\)

AC=CE = 2x

Also,
\(\frac{AB}{AD} = \frac{BC}{ED}\)

We get BC = x = EF

AC+CE+EF = AF

\(5x = \sqrt{20^2+10^2}\)

\(2x = 4\sqrt{5}\)

\(CE^2=(2x)^2 = 16*5=80\)



When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
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\(\frac{AD}{AB}= \frac{AE}{AC} =k\)

\(\frac{BD}{CE} = \frac{AD-AB}{AE-AC} = \frac{AB*k-AB}{ AC*k - AC} = \frac{AB(k-1)}{AC(k-1)} = \frac{AB}{AC }\)

\(\frac{BD}{CE}= \frac{AB}{AC }\)

\(\frac{AB}{BD} = \frac{AC}{CE}\)

OR we could apply Dividendo (Ratios property) to prove it.

OR The Triangle Proportionality Theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.





SiddharthR


When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
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nick1816
\(\frac{AD}{AB}= \frac{AE}{AC} =k\)

\(\frac{BD}{CE} = \frac{AD-AB}{AE-AC} = \frac{AB*k-AB}{ AC*k - AC} = \frac{AB(k-1)}{AC(k-1)} = \frac{AB}{AC }\)

\(\frac{BD}{CE}= \frac{AB}{AC }\)

\(\frac{AB}{BD} = \frac{AC}{CE}\)

OR we could apply Dividendo (Ratios property) to prove it.

OR The Triangle Proportionality Theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.





SiddharthR


When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.



That's interesting. Thank you for that insight. Could you please also tell me how you arrived at length "2x" for AC & CE ?

I got the following - AB/AD = 10/20 = 1/2;
Let BC = x. Then BC/DE = 1/2; From this we get DE = 2x

From here, I am unable to understand how you reached 2x distance for AC & CE.
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