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In Fig. $6.20, \mathrm{DE} \| \mathrm{OQ}$ and $\mathrm{DF} \| \mathrm{OR}$. Show that EF IIQR.

Geometry

Chapter 6

Triangles

Section 2

$\quad$ Similar Figures

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we are going to do problem number five in this question. Uh The given things are first uh D E. Yes, parallel to OQ. And another thing which is given is D. F. Is parallel to O. R. So we need to prove to prove that is E. F. Is Berlin to cure. This is the given figure. And if you see given that Eddie is parallel to this one Cure. And this D. F. Is parallel to the O. R. So let's just go through this. Mm. So right here proof. No. First of all, we will go through in uh triangle speak you. Oh okay. So if we just go through triangle PQ. Oh and it has given that D. E. Is parallel to Tokyo means that line the will cut this cross burning sorts of side in a proportion according to the cure. So we are going to destroy it in proportion. So this is a P. E. Divided by EQ. This will be equals to feeding divided by an audio. So what does this take this as a question one. Now we will talk about in triangles. C. O. R. No, again, it is given that D. F. Is the line which is parallel to you are the simply means DFL cut the two sides of the triangle in a proportion. So you'll be writing this as speedy divided by D. O. This is equals two pdf by F. R. So we will be taking this equation number two. Now, if you just compare equation on an equation to so one side that is a pretty wide deal pretty but you're equal there. So other side of the two sides will be also called PE by EQ concept here by fr So right here from here later. So P. E. Bye. E. Que. This will be equals two pdf Bye Airport. Okay, No, uh if you see this part is uh sewing uh and if you consider it a triangle peak, you are okay then lying E. F. Is cutting two sides of a triangle in a proportion. Okay? As it is given PE by EQ. Calls appear by a far as it is cutting it in proportion. So the line ef we panel to CUBA responded to your um So as but Children as part of your um Line E. F. Will be parallel to CUBA. Okay? So this is proved now. So this is the answer. Thank you.

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