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 [imath]HK=\frac{1}{4}c[/imath] and [imath]FE=\frac{1}{2}c[/imath]. Since it was proven that the two triangles are similar I got that:
[math]\frac{XH}{XE}=\frac{HK}{FE}[/math][math]\frac{XH}{XE}=\frac{\frac{1}{2}c}{\frac{1}{4}c}=2[/math]
But I am not sure on how to progress from here.
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 [imath]HK=\frac{1}{4}c[/imath] and [imath]FE=\frac{1}{2}c[/imath]. Since it was proven that the two triangles are similar I got that:
[math]\frac{XH}{XE}=\frac{HK}{FE}[/math][math]\frac{XH}{XE}=\frac{\frac{1}{2}c}{\frac{1}{4}c}=2[/math]
But I am not sure on how to progress from here.