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In Fig. $6.39, \mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively. Prove that:(i) $\Delta \mathrm{ABC}-\Delta \mathrm{AMP}$(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
Geometry
Chapter 6
Triangles
Section 3
Similarity of Triangles
Similarity
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we are going to do problem number nine. And discussion, we have to just go through the given figure. And in first product we have to just prove that triangle abc is similar to strangle A. M. P. Okay, so if you see in this case that this angle C. Will be common and this angle M. And this angle B. They both are 90 degree. So you can simply write it as angle C equal strangle. See that is common in this two triangle. Now, angle M is equals two angle B. This is 90 degree H. Okay, so by using a similarity, great area, a similarity cafeteria these two triangles will be similar. That triangle abc will be similar to triangle A. M. P. Okay, so this is the proof of first part. No, let's just do the second part. In second part. We need to prove that C. A. By P. A. This is equals two B. C by M. B. Okay, No, as uh as we have just proved here, just right now, we have just proved that triangle abc Okay. Similar to strangle A. Mp. Okay, By a similarity criteria. So if to uh, triangles are similar then by using the property of basic personality cure. Um you can see that's corresponding sides will be also equal. That is we can write that C. A. By P. Let them just so you know, diagram here. This is uh see A by this one that is P. A. Okay, this should be equals two B. C. By mp. So this is A. B. C. And this is empty. So the corresponding surge will be equal. So we can see A by P. And these two vehicles too busy by empty. This is the proof the only thing which we are just toys. And these two triangles are similar and we have just used basic proportionality theorem. Okay, that is Children. 6.1. Just try to clearly this is europe. So this is uh the answer. Thank you very much.
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