 Hi. I'm Zor. Welcome to Unizor education. This is second lecture of problem solving about triangles, quadrangles, stuff like this. There are, I think, eight lectures all together dedicated exclusively to problem solving. I do encourage everybody to look at these problems and try to solve them yourself. It's extremely useful and it actually builds up your creativity, which is not even related to mathematics, but it's related very much to my most important point that mathematics really develops your creativity, which you can use anywhere, any other industry, profession, whatever. Let's just do it. Prove that the medium like hypotenuse in the right triangle is equal to half of the hypotenuse. Okay, so you have the right triangle. Now, you have to prove that the medium to hypotenuse is equal to its half. Well, let's think about it this way. Since it's medium, we know this is equal to this, right? Now, if I will extend the medium by the same size and build a quadrangle, now this quadrangle will be obviously a rectangle. Well, first of all, it's a parallelogram, because since these two are equal and these two are equal, so these angles are vertical. That makes this line, this segment, equal to this segment from triangles A, B, C, D. So A, B, P triangle is congruent to D, P, C by side equal to side, another side equal to this side and the vertical angle in between. That makes A, B congruent to C, D similarly. B, D is congruent to A, C. So it's a parallelogram and it's not just a parallelogram. This angle is right angle. And since interior angles of parallelogram are either supplemental or congruent to each other, that makes all angles right. So every angle is 90 degrees, so this is a rectangle and since it's a rectangle, its diagonals are congruent to each other. But one diagonal is a hypotenuse, the other is double-needian. If they are equal to each other, that means that the median is equal to half of a hypotenuse. Proof-converse theorem. If median in a triangle is equal to half the side, it falls on when the angle where it starts is the right angle. Okay, so let's consider we have a triangle and it's equal, it's half, half of this side. So these are three equal segments. Well, how to prove that this angle is the right angle? Well, actually there are many ways of doing it. Here is, I would say, an unusual way. Well, the usual way is to exactly the same thing. Extend it, have a parallelogram, and prove it from the parallelogram that this is the right angle. But here's something which just came to my mind. I think it's quite interesting. You see, these are congruent segments, so are these that makes this angle equal to this one and this one equal to this one, because these are isosceles triangles. So A, B, C, D. A, D, B is isosceles, since median is equal to half of the side it falls on. And this makes an isosceles triangle. So it looks like some of these two angles is equal to some of these two angles. But what do we know about the angles in a triangle? That some of them is equal to 180 degree. So this, plus this, plus this, plus this is 180. But this is half of this, since piece of this angle is equal to this one and another piece is equal to this one, so all together that makes it half of 180, which is 90 degrees. I think it's more interesting than with parallelogram. But as I was saying, there are many different ways to prove the same theory, so this is just yet another way. By the way, the previous lecture was mostly for construction problems. These are mostly problems related to proving something. Proved it in the right triangle, so median and an altitude to a k-potronus form an angle equal to difference between the triangle's acute angles. Okay, so you have right triangle. Okay, now you have a median and you have an altitude. Ag is median, Ae is altitude. Now, the angle between them is supposed to be equal to the difference between two acute angles. Well, let's do it this way. Since this acute angle is equal to this acute angle, right, since we know that EG, median, is equal to half of a k-potronus, Ae is also this triangle, and so is Dac. So this angle is also the same. Like this. Now, what else is interesting? Ae and B and Ec and Ac. Aec. Aec also the right triangle. So what's the interesting point about Aec and ABC triangle? Well, both are right triangles, right? Ac is altitude, so this is the right angle, and in ABC the A angle is right. But what's important is they share the same acute angle, which means the other acute angle is the same. So this other acute angle, Aec is congruent to ABC. Again, since both ABC and ABC are right triangles, which share the same acute angle, another acute angle will be the same, will be congruent. And now it's obvious that angle Dae is a difference between Dac, which is equal to this one acute angle, and Eac, this one, which is equal to another acute angle. So the difference between the bigger and smaller acute angles gives you the difference between the median and an altitude of the right triangle. That's it. Okay, given triangle ABC segment, AG bisects angle Bac. ABC and bisector. So AG is a bisector. Point D lies on BC, okay. Straight line through point D is parallel to side AC. So we have a parallel line here. And then through E parallel to BC intersecting AC at point F. These lines are parallel and these lines are parallel. Prove the segments Ae and Cf. Ae and Cf are congruent. Well, let's think about it. Well, since these lines are parallel, these are lines are parallel, E, G, C, F is a parallelogram. So this guy is equal to this one. Now, since AG and AF are parallel and AG is transversal, these are alternate interior angles. Ae, E, G and AF parallel and AG is transversal. So these are alternate interior angles. Now, since AG is bisector, these two angles are equal to each other. And that's why these two angles are equal to each other. E, A, D and E, D, A. And that's what makes triangle Ae, D isosceles. Since angles at the base are equal to each other, the sides are congruent. And that's what makes E, D congruent to E, A. So FC is congruent to E, D because it's a parallelogram and E, D is congruent to AE because AE, D is a sosceles triangle. That's what makes FC and AE congruent. That was easy. Okay, given an angle, MXN. Inside it an angle P, Q, Y. So there is one angle and there is another angle. In such a way that MX, so MX is parallel to PY, MX, MX is parallel to PY. And what's important, the distance between MX and PY between MX and PY is equal to the distance between these two. So since distance is measured by perpendicular, we can say that if you take the point and drop the perpendicular here and there they will be equal to each other. So the distance is the same. Parallel and parallel. Distance between parallel lines here and there is the same. Now, since the distance is measured by a mutual perpendicular so I used the Y point and dropped to perpendicular to both sides of MXN to get these two called A and B. Now, that's all the condition. Proof that bisector of angle MXY is a bisector of PY, Q. So they share bisector. That's what's necessary to prove. Okay, first of all, bisector of angle MXB must actually contain this point, Y. I'll just prove that these two angles are congruent to each other. Now, these are two right triangles, X, A, Y and X, B, Y with congruent legs because the distance between parallel lines is the same and common hypotenuse. So that's why two angles are congruent to each other which means the bisector goes to Y, to point Y. Now, what if I continue this bisector? How can I prove that these two angles are also the same? Well, that's very easy because this line and this line are parallel and X, Y and its continuation is transversal and angle MXY and angle PY whatever W are, how is it called? Corresponding. Yes, corresponding. These are corresponding angles. So PYW and MXY are corresponding with AM and PY parallel and XYW transversal. Now, same are these two angles W, Y, Q and Y, X, B also corresponding angles. And they are equal. So this is equal to this, this is equal to this. These guys are equal among themselves. So these are equal. So the continuation of line XYW is bisecting the angle PYQ and the proof. Proof that any segment that connects to basis of a trapezoid is divided by a median in two congruent parts. Any segment that connects to basis of a trapezoid. Okay, so you have a trapezoid, you have its median. What's necessary to prove is that any segment connecting two points on basis is divided in half by a median. So we have to prove that MX is equal to XN. Well, what do we usually do in cases like this? Well, we make a parallel shift towards one of the vertices. So let's call M was the original letter and this will be M' So if we shift MN to the right so that M is gliding towards C and M would be gliding towards D into a new position M, M and prime. Now what happens? Well, it's a parallel shift. So considering lines BC and AD are parallel while M will be gliding along BC towards C M will be gliding along AD towards M prime position. And obviously M and CM prime is parallelogram. And what's interesting is that PQ, which is parallel to BC and AD will be dividing the triangle M'CG into pieces and this point would be half. This would be X prime. Now why is it half? There was a theorem about triangles that if the line parallel to a base crosses the side in the middle it will cross another side in the middle as well. So this is the line PQ parallel to the base and it's crossing one side at point Q in its middle. That's why X prime would be in the middle of the CM prime. But now let's consider M and CM prime. This is parallelogram, right? And since PQ is parallel to both then it's two different parallelograms actually equal to each other. X and C X prime and MX X prime M prime are both parallelograms which means that this side is equal to this this side is equal to this but since C X prime is the same as M prime X prime so they are congruent to each other that's why these also will be congruent to each other. And that's what proves that XM and MX have the same lengths. Again, this and this, these two segments are congruent to these two and these two are congruent among themselves from the triangle CM prime P. That's it. Given the triangle ABC let's vertex B be on the top and side AC be at base. Point X is an intersection of two bisectors of angles at the base. Okay, so you have one bisector and another bisector and crossing is X. Okay, straight line through point X parallel to the base intersects at points M and M. Okay, so these are bisectors and MN parallel to AC. Proof that segment MN is equal to a sum of segment AM and NC. Okay, well that's actually, again, such a long condition of this theorem but the theorem is actually trivial. Here is why. MN is parallel to AC. AX is transversal so these two angles MXA and XAC are alternate interior and that's why they are congruent to each other but at the same time since this AX is a bisector these triangles are congruent so that's what makes these two congruent and that's what makes AMX a nice socialist triangle with these two segments equal to each other. Same thing on this side. Since XM is parallel to AC XC is transversal so NXC and XCA are alternate interior angles and congruent to each other. Now, since XC is a bisector these triangles are also congruent to each other so these are congruent and these are congruent that's what makes these two congruent that's what makes these two sides XM and NC congruent to each other and that's what makes MN equal to sum of AM because it's equal to this piece and NC which is equal to this piece. Straight lines are drawn through all three vertices of a triangle forming another bigger triangle. Okay, so we have a triangle and we have three sides three lines parallel to something like this. This is our original triangle, ABC and this is our new triangle let's say MNP So, what's necessary to prove here? We look the same, right? Okay, prove that this bigger triangle is divided by size of a smaller triangle into four triangles each congruent to the small one and side being twice as big as the parallel side. Okay, so what do we know? We know the parallelism, right? Now, since this is parallel to this and this is parallel to this that makes AD and C parallelogram, right? Which means this and this are congruent. This and this are congruent. Now, ABCP is also parallelogram which makes this line congruent to this line. Now, NBCA is also parallelogram so this is the same as this. Now, MACB is parallelogram and AC and MB are congruent. And finally, missing what? ABCP is parallelogram so this side and this are congruent. So what do we see? Well, basically, we have already proven everything. First of all, all these four triangles are congruent among themselves by three sides, that's number one. Number two, what we see is that the length of this guy is half of this guy. So big triangle is twice as big as the small triangle and each side is twice as big as the one of the small one which is parallel to the big one. That's what's necessary to prove. Again, to draw and to explain what we want to do is longer than the proof itself, proof is trivial. Prove that in a socialist triangle, sum of two distances from any point on the base to two legs is constant and equal to an altitude. Okay, that seems to be a lot of easy truth. So if you have a socialist triangle and you take any point here and distance to this and distance to this, two perpendiculars. So this theorem states that sum of these two is constant which does not depend on the position of this point and is the same as an altitude towards the leg. So basically if you move this point to the right, now this piece will be longer, this piece will be shorter, but the sum of them will be exactly equal to this one. That's what's necessary to prove. All right, so how can we prove this? Well, let's do this way. What if I will draw a line parallel to the side to the leg like this? So these are two parallel, which makes this piece is equal to this one. What I have to prove right now is, so, one more. From point M I draw a parallel and E to the leg AB. Now, this is a socialist triangle, don't forget that. Now, since these are parallel lines, I can say that angle BAM is congruent to EMC because these lines are parallel and AC is transversal. So these are two corresponding angles. And since BAC is equal to BCA, that's why angle AEMC is equal to ECM. And obviously it's equal to this one. All right, so this is obvious. These are corresponding with this parallel and this transversal. Now, since this angle is equal to this one in a socialist triangle, so I have this is equal to this two angles are also congruent to each other and that's why EMC is also a nice socialist triangle. So whenever I cut piece of a nice socialist triangle by a line parallel to a leg, I have left also a socialist triangle. Now, here is what's interesting. From my altitude Cd, this is the piece Yd which is equal to the Pm. So this distance from point M to AB, which is MP, MP is perpendicular, is the same as the distance from point Y to D. If I will be able to prove that this piece, CY is equal to MQ, then my theorem will be proven completely because now I can say that MQ plus MQ is equal to DY plus YC, right, which is the full altitude. Now, that's actually much easier to prove that XC is congruent to MQ because we are dealing with a socialist triangle. Just forget about this part. Just consider only the socialist triangle. What we do have here is two altitudes and two different legs, which are obviously congruent to each other. Well, number one, because we have already considered this problem before, but number two, it's really easy since these are right triangles, this one and this one, which share a part in those and the same acute angles since it's a socialist triangle. So these two altitudes are the same. That's why MQ and CY are the same. But now MQ, we just add a piece MP and that's what the distance to another side and similar piece YG being added to CY gives us the full altitude. So no matter where point M is located, some of these two distances is equal to the altitude to the side. Okay, I have the last problem. Change the condition of the previous theorem to use a point on continuation of the base outside of the triangle and formulate theorem in this case and prove it. That's interesting. So you have also a socialist triangle, but now instead of being inside the triangle, point M is outside. Now distance to this and distance to this. Most likely it's the difference between these two distances which would make an altitude. If M is inside, that's the sum. But if M is outside of the triangle, then probably the difference between the distances of two legs of a socialist triangle will make an altitude to the other. Now how to prove it? Well, the same way. Let's do it this way. We will draw a parallel line to AB through C. Now since M, M, P, Q. Okay, since M, M is perpendicular to AB and C, W is parallel to AB by construction, M, Q is perpendicular. Now this angle is equal to this one as corresponding angles with these parallel and transversal. Now this angle is equal to this as an angle in a socialist triangle at the base, and this angle is equal to this as vertical, which makes these two angles Q, C, M and M, C, R equal to each other, which makes triangles C, Q, M and C, R, M congruent because they are right triangles. This is the perpendicular, don't forget. And this is the perpendicular. So these are right triangles with common hypotenuse and an acute angle. And that makes these two segments Q, M and R, M congruent to each other. And now you see that the altitude P, C which is equal to MQ obviously because these are parallel and perpendicular. So the difference between M, M and R, M is equal to MQ since R, M and Q, M are the same. And Q is in turn equal to PC. So that's why we can state that the difference between the distances from point M outside of it is also a triangle on its base. To both legs is equal to an altitude to a leg. That's the end of it. Okay, that's it. That was the last problem to this lecture. Don't forget Unisor.com has this and many other lectures as well and for parents and supervisors it provides an excellent opportunity to control and supervise the educational process of their children and students. That's it. Thank you very much and good luck with other problems.