Ratio Proof

[Tarmac27]

Hello,

I am having trouble on the last part c)iii) on this geometry question. I have included the other parts just to give context.


From c)i) I found that $HK=\frac{1}{4}c$ and $FE=\frac{1}{2}c$. Since it was proven that the two triangles are similar I got that:
$$\frac{XH}{XE}=\frac{HK}{FE}$$$$\frac{XH}{XE}=\frac{\frac{1}{2}c}{\frac{1}{4}c}=2$$
But I am not sure on how to progress from here.

 

[Steven G]

Please show your work as required by the posting guidelines.
Show why HK = .25c and FE = .5c
Which triangles are similar and why are they similar.?

 

[Tarmac27]

Please show your work as required by the posting guidelines.
Show why HK = .25c and FE = .5c
Which triangles are similar and why are they similar.?

$$HK = -\frac{1}{2}OE+OC+\frac{1}{2}CF$$$$HK=-\frac{1}{2}(\frac{1}{2}a+\frac{1}{2}c)+c+\frac{1}{2}(\frac{1}{2}a-c)$$$$HK = \frac{1}{4}c$$
$$FE=\frac{1}{2}OA+\frac{1}{2}AC$$$$FE= \frac{1}{2}a+\frac{1}{2}(c-a)$$$$FE=\frac{1}{2}c$$
Triangles XHK and XFE are similar by SSS test since all the corrosponding sides are parallel only different sizes

 

[lev888]

$$HK = -\frac{1}{2}OE+OC+\frac{1}{2}CF$$$$HK=-\frac{1}{2}(\frac{1}{2}a+\frac{1}{2}c)+c+\frac{1}{2}(\frac{1}{2}a-c)$$$$HK = \frac{1}{4}c$$
$$FE=\frac{1}{2}OA+\frac{1}{2}AC$$$$FE= \frac{1}{2}a+\frac{1}{2}(c-a)$$$$FE=\frac{1}{2}c$$
Triangles XHK and XFE are similar by SSS test since all the corrosponding sides are parallel only different sizes

SSS does not say triangles are similar if sides are parallel.