Hình 9A1 Bài 1

Half of the circle tangent A-X Take point C on the circle On the circle Ray B-C Intersects ray A-X At point D Tia A-C is cut by tia B-Y at point E Call M the midpoint of A-D M is the midpoint of A-D Which is the midpoint of B -E Question 1 Prove that M-L is a tangent line We need to prove that M-L is a tangent line Question 2 The C-H A is the root of A-B Prove that A-L A-L C-H is correct according to the rules And let's observe like this And we make the most superficial inference like this First of all We have a right angle BI is a tangent At the same time we also have angle C as a right angle because the angle that looks at the meridian is a right angle We also have angle C as a right angle I examine triangle A-C-D which is right-angled at C I examine triangle A-C-D Now this angle is the right angle Again C-M is the tangent Because M is the midpoint of A- D Therefore C-M is the tangent of this right triangle Now I have a theorem in a right triangle the median to the hypotenuse is equal to half the hypotenuse A-C-D Midline C-M A flame next to the hypotenuse So we have C-M right away By M-A By M-D Because of the symmetry of the figure we also obtain N-C N-C N-B So we have these three people who are equal Now let's prove that M-N is a tangent How to prove that a line is a tangent Do we need to prove that This O-C angle is right-angled with M-N isn't it We are proving that this angle is right-angled MN is tangent if and only if OC is perpendicular to MN That is we need to prove that OC is perpendicular How do we usually prove that angle is a right angle How do we usually prove that angle C is a right angle Is there a lot of ways to prove a right angle Usually we have the standard method for this angle For example we prove that these two angles add up to 90 degrees This method of proving the angle is quite common There is another way to prove it using another right angle That is we prove that this angle C is a right angle by using another right angle So the right angle we just mentioned has angle A angle C and angle B right So here let's see if this angle C can be equal to another right angle What do we do to prove that two angles are right angles What do we do to prove that two angles are right angles That's how we will prove that they are two angles of two equal triangles right Two angles of two equal triangles Then in this diagram we can immediately see that the MC is equal to MA and AO is equal to CO because they are also radii of the circle right Monday is the MO which is common Therefore the two triangles are MAO and triangle MCO which are equal In the case of side-side-side triangles then this angle C is equal to angle A And therefore this angle C becomes a right angle because angle A is right Here we have completed the verbal solution for the proof by linear solution How can Kim now repeat the solution First we have a right angle at C because it is viewed from a vertical angle at a right angle Then triangle BCA is a right triangle at C We have M is the midpoint of AD That means the CM is the median of the right triangle So the MC is half of BA and therefore this MC is by MA Thus the triangles MAO and MCO are congruent by the side-side-side case Then I will lead to the right angle C And therefore when angle C is a right angle I have a linear solution Now to the second floor Here we go let's get started How can I prove that three lines are concurrent One of the more common methods is to prove that it is three lines in a triangle For example if it's the three medians in a triangle they will intersect right Or we can prove that they are the three altitudes in a triangle and they are concurrent or they are the three angle bisectors Then that is also a quite common way However by observing the diagram we can see these three lines: AN BM CH It is the third line within the triangle ABC Then CH is the altitude of triangle ABC But the BM and the CN are not the same In conclusion the path to prove that it is three special lines in a triangle is not We didn't succeed so we won't take that path anymore That's where we'll find another approach Now pay attention it's our CH Our CH is it's in the corner with AB So that means it will be parallel to AD and parallel to BE right CH is perpendicular to AB Then CH will be parallel to AD and also parallel to BE Is that right Then how do we observe a situation like this If we now call you K the knife of MB and CH We call K the knife of MB and CH K is the knife of MB and CH Then something like this happens There is a result of a trapezoid If a trapezoid ABCD is given with AB parallel to CD then the midpoint N of CD and the midpoint M of AB The intersection point Y of AD and BC Then Y M N will be collinear Or it is called YM will pass through N Either YN passes through M Or YM is collinear Where M and N are the midpoints of AB and CD This proves to be very simple We use Thales' theorem and can prove it immediately Therefore this is a result of a trapezoid Then it will lead to something like this I consider the CD app as a trapezoid Then the two sides AH and CD are different on this side M is the midpoint of AD Then MB cuts CH at point K Then K will also be the midpoint of CH Then it's very simple Theo tỷ lệ thì CK chia cho MD bằng KH chia cho MA vì cùng bằng BK chia BM đúng không Then CK divided by MD equals KH divided by MA But MD and MA are equal Then it proves that CK and KH are equal It means that if MB intersects CH at a point K then point K will be the midpoint of CH In other words MB passes through the midpoint of CH So we have that This is called the MB and CH knife Then we can deduce that K will be the midpoint K will be the midpoint of CH In other words MB passes through the midpoint of CH right Then now we switch the positions AN is also completely similar in that way Then AN also passes through the midpoint of CH It's completely the same We can also say that AN passes through the midpoint of CH which is point K So we immediately get AN and BM Both pass through the midpoint K of CH So that means these three lines are concurrent at K right We are conducting the verification process The beauty of this solution is to prove that three lines are concurrent The proofs show that these three lines pass through one point To prove that three lines are concurrent we prove that they pass through the same point These two roads pass through the same point Of course the problem of proving three concurrent lines is also a very interesting problem and there are many different ways to think about it For example directly we will examine Problem 2 Problem 2 is also a problem proving that three lines are concurrent And we also have a very good direction To develop multidimensional thinking skills We can ask a question such as Are there other approaches to this situation For example this one here For example I call K the knife of M B and N A And we prove that we need to prove tha We need to prove that CK CK passes through H If the CK in the corner AB is equal to then CK will coincide with AH Now we want to prove the angle Prove that CK is parallel to AD To prove CH is concurrent with AN and MB then we call K the intersection point of AN and MB We will prove that CKH are collinear But the CH We have already proven that AB is perpendicular to the circle Then we prove that CK is also perpendicular to AB With AB But to prove CK is perpendicular to AB Then we prove that CK is parallel to AD because AD and AB are perpendicular to each other meaning we prove that CK is parallel to AD Now let's temporarily consider the perception Then to prove parallelism of CK with MA we will have to prove that the ratio of NC divided by MC is equal to the ratio of NK divided by Where are you going We can write it like this Parallel CK With MA When and only when we must prove that CN is divisible by CM is divisible by KA Is that right But something very nice happened Your CN Then it turned out to be NB Because NC and NB are the gland NC equals NB and MC equals MA So this CN will be replaced by NB and this CM will be replaced by MA Then I will do it Then what is it now What will it be equivalent to Equivalent to NB Equivalent to NB divided by MA My CN is replaced by this NB CM is replaced by this MA Now I have to prove that the two ratios are equal KN On KA Then look over here Who is NB on MA NB on MA Is there a KN on KA Then because the two of them have already been parallel Therefore the matter This person divides this person This is the same as theorem of the deaf ear It is theorem of the deaf ear Therefore this is correct What does this correct mea It means we have proven the success of parallel CK with MA Then the teacher thought that this is a second very good direction If now If I were to present it again in a longer way How would I present Teacher is repeating again Let KA be the segment of AN and BM We have it CN is divided into CM by the number Note Chia MA Because CN is equal to NB And CM is equal to MA But if NB divided by MA is equal to NK divided by NA then according to the calculation CN divided by CM is equal to KN divided by KA and thus Check the triangle MAN is divided by CN into CM By KN divided by KA Then according to theorem we have CK parallel to MA But MA is parallel to AB Therefore I also have CK parallel to AB Oh right CK is perpendicular to AB And we have CH also perpendicular to AB CK is the same as CH Thus CH will pass through K This means that AN BM and CH The button K Then we successfully prove the second method