 Hi, I'm Zor. Welcome to Unizor Education. I continue my favorite subject, Problems, related to similarity in this case. This is lecture number 5 with different problems. You definitely have to try to solve all these problems by yourself first, before listening to this lecture. So I hope everything is done. Whatever you have been able to solve, great. If you didn't solve some problem, it's OK too. I will do right now all these problems in front of you. And when I finish, it's very important to basically repeat just by yourself all these proofs which I have presented. And that would be the great way to kind of put your logic in place so it will be useful for some other purposes. What kind of problems are we facing right now? Given a circle with a diameter AB and quart PQ perpendicular to the diameter. So there is a circle, this is diameter AB which goes through the center and quart PQ perpendicular to it. Intersecting at point X. Prove the following equality between the lengths of corresponding segments. 4 times AX times XV is equal to PQ square. OK, how can I prove that? Well, you remember that angle supported by a diameter is the right angle, right? It's basically half of the angle which is 100 degrees, which means it's 90. Now also, since I said this is perpendicular, this is also the right angle. Now let's consider two triangles APX and BPX. Now obviously these two angles are congruent to each other because this is perpendicular to this and this is perpendicular to that. Which means that these two triangles are similar. Now from similarity of these triangles we conclude that the catatose which is against this angle in a bigger triangle which is PX relates to the catatose which is opposite to the same angle in a small triangle which is XB as they are related to the other pair of catatose which are lying opposite to other two angles which is AX to and in a small one it would be PX. Now from which we conclude that AX times XB is equal to PX square. Now PX is one half of PQ because as you know the radius perpendicular to a chord divides it in half. Again, that was one of the problems which we were solving before. Many theorems or whatever it is. So PX is actually one half of PQ square which is equal to one fourth of PQ square. From which this four goes here and you have the PQ square is equal to four times AX times XB. What's supposed to be pretty good. Ok, that's simple. Now notice that in this and in all other problems related to similarity all I have to do is to find or construct some couple of triangles usually the right triangles where some angle is common or congruent which means they are similar and then just use the proportionality of the sides. Basically this is a typical example. Just find something which is similar and use the proportionality. Now let's go next. Given a circle with diameter AB starts the same way and chord intersecting this diameter PQ intersecting at point X. Prove the following equality between the lengths of the corresponding segments. AX times XB equals PX by XQ. So AX times XB equal to AX times XQ. Now I'm not saying that the chord is perpendicular in this case it's different and that's why these two segments have different lengths because if it was perpendicular the chord would be divided in half by the bandwidth and now it's not. However this particular equality still holds. Now how can we prove this particular thing? Well let's just connect these two things. Now what do we see right now? You know that inscribed angle is measured, inscribed into a circle I mean, is measured as half of the central angle which is supported by the same arc. So angle ABP is measured as half of the central angle supported by this arc and so is angle AQP. So they are congruent to each other because they are supported by the same arc. Similarly these two angles are congruent because they are both supported by this arc VQ. Now obviously we have two different triangles with two corresponding congruent angles which means they are similar. So again we found the similarity and now just use this similarity to prove this particular equation because this is definitely just a proportion. So the side in the bigger triangle across this angle is AX. This side in the smaller triangle which is opposite to the same angle is XE. Equals. Now let's take another pair of angles. This and this. In this triangle opposite to this angle is XQ and in this triangle opposite to this same angle is XB. So AX times XB, AX times XB equals to PX, XP, PX, same thing, times XQ. All these just found the similarity between certain triangles. Well actually I can't say that it was absolutely trivial because you see if you have a diameter in the chord you don't have triangles, right? So I had to draw a couple of lines from the ends of diameter to ends of chord to basically construct a couple of triangles and then I proved them to be congruent. Sorry, similar. So just some small very very small and kind of intuitively obvious additional construction I had to make before going into the conclusions. Okay, next. Given a circle and point X inside it let AB be any chord that contains this point. Prove that the product of lengths of segments AX, AX and XB is independent of the position of the chord AB as long as it contains the point X. So we have to prove that AX times XB is constant. Regardless of position of the chord. This way or this way or this way. So AX times XB is the same as A prime X times XB prime same as X double prime X to XB double prime etc. Now this is something which is an immediate consequence from the previous theorem because if I will draw a diameter through point X and call it mm. Then we know from the previous theorem that the multiplication of the lengths of this times length of this is equal to result of the multiplication of this times this. That was a previous theorem. And again it doesn't really depend on whether it's this chord or this chord or that chord. In any case this is all equal to mx times xm where mm is a diameter which is going through this point X. That's why this is a constant regardless of the position of the chord as long as the point X through which the chord actually goes is fixed inside the circle. Direct consequence from the previous theorem. Actually it doesn't really deserve to be a separate problem. I would say it's just a continuation of the previous problem. Okay, what's next? Given a circle and point outside it, let xm be a tangent. Tangent as you know is always perpendicular to the radius to a point of tangency. Okay consider any line that contains point X and intersects a circle at two points a and b. And prove that the product of lengths of segments xa and xb xa times xb is equal to square over distance from x to m. Okay, for obvious reason we have to find again similar triangles and again something which is intuitively obvious we have to connect this to this and this to this. So I wonder if it is obvious that triangles xbm and xam are similar to each other. Well, maybe it's not so obvious but anyway they do have the common angle, right? This one. Now, how about the other end? We have to prove that this angle is congruent to this one. Well, number one, there was a theorem which I have already proven before that if you have a circle tangent and a cord let me draw it more or less the same way here, tangent and a cord, then this angle is measured as half of a central angle supported by the same cord. Now, I'm just referring to this theorem and now using this theorem I can say that, okay, since this angle is equal to half of the central angle supported by this particular cord and this is also this is an inscribed angle which is also equal to a central angle, half of the central angle supported by the same cord. That's why these two angles are congruent. I do refer you to this particular theorem. That was in a lecture when I was talking about inscribed angles and angles formed by cord and the tangent. It's all in that lecture if you don't remember exactly how to prove it I would suggest you to try to prove it yourself first and then if you can't go to the lecture, listen to the lecture again, it's very important for you to be able to prove it basically. It will, so to speak based on whatever the experience you have already accumulated in proving different theorems. So I'm using this particular theorem. So this angle is equal to half of this arc, half of the central angle associated with arc, as well as this angle. So they are equal to each other which means that the triangles are similar. And from similarity of the triangles I'm sure we can get this one. Alright, in a small triangle let's go small to the big one. In a small triangle let's take the side which is opposite to this double arc, which is XB. Now in a bigger triangle the side opposite to the same angle is in XMA it's XM. Now in a small triangle XBM the side which is opposite to the third angle, not this, not that, but the third angle, this one which is same as this one, not this one, is XM. And in a bigger triangle the side which is opposite to XMA is XA. From which we have XB times XA, XA times XB is equal to XM square. So all we have to do is to connect these two points and again prove the similarity of the triangles which is, well, it's intuitively obvious but obviously we have to prove it. And the way to prove it is one angle is common and another angle, pair of angles, are congruent to each other because of certain theorem which we have already proven before. Next, given a circle and point X outside it, consider any line that contains point X intersects a circle in points A and B. Prove that the product of length of segment XA times XB is constant as long as I mean constant regardless of the direction of this line as long as it goes through X and intersects our circle in two points. And again, this is actually a direct consequence from the previous theorem because if I will draw a tangent the product XA times XB would be a square of a distance from X to the point of tangency. So regardless of direction of the XA this way or this way the product of bigger times smaller lengths of these segments is always equal to square of the segment within the tangent and that's why it's constant. And by the way if I will take another tangent, these two distances from XM to XM are equal to square and that was again a different theorem which we have proven when we were talking about circles and tangents etc. That's why it doesn't really matter which tangent I will take one or another. This is always square of that piece of tangent. Well, that's it for today and I do recommend you to go through all these problems again. Go to the website unisor.com and listen to these problems. Try to prove all these equations yourself and if you're comfortable great. If not try to listen to lecture again. Website is open for all. Teachers and parents can use it for home schooling or group study and whatever else. That's it. Thanks very much.