Hình 8A1 Bài tập số 1 #toan8

The problem involves square ABCD with point M moving along BC moving along the flat BC and flat BC and a perpendicular AN is drawn to AM where N lies on the line CD Some people call it the line from A that is perpendicular to AM and intersects the line CD at point N Let H be the midpoint of N then HM and HL are equal and from M draw a straight line parallel to AB the line AK intersects the line AH at point E and the line AH extended intersects CD at point K They asked me to prove three things the first is to prove that AM equals AN To prove that two line segments are equal it is common to introduce two equal sides of two triangles Then on this figure we have the temporary angle ABM and the temporary angle ADN en we immediately have AB and AD equal these are two sides of a square The second is that angles B and D are equal because D are both right angles The third condition is that angle A1 is equal to angle A2 which we can see simply We have angle A2 plus angle A3 equal to this angle A which is a right angle right The angle A1 plus this angle A3 is also a right angle That's why people say that A1 and A2 are equal in the sense that these two angles are both complementary to the same angle That is it is equal to 90 degrees minus this angle A3 In conclusion we obtain that the temporary triangle ABM and the temporary triangle ADN are equal In the case of angle-side-angle this angle B is equal to this angle D side AB is equal to this side AD angle A1 is equal to this angle A2 this is angle-side-angle And thus we obtain what was to be proved that AM is equal to AN Second prove that EMKN is a parallelogram a rhombus Usually to prove that a figure is a rhombus one would first prove that it is a parallelogram We'll start by proving the parallelogram So by placing two equal triangles triangle EHM and triangle KHN Then these two temporary peaks are clearly visible here with two H-shaped peaks here The second thing is that HM and HN are equal So now we have a pair of equal angles and a pair of equal sides so we naturally look at these angles M and N Then we immediately see that it is solid because there is a situation like this EM assigns the problem in parallel AB but AB is also parallel to CD So I have the EM CD in parallel And therefore angle M and angle N are equal as they are two solid angles So in short we have obtained a temporary MHE by using a temporary NHK In the case where the angles at the corner are equal angle H is equal to angle H side HM is equal to side HN and angle M is equal to angle N Because the two triangles are equal we get that the corresponding sides EM and NHK are also equal EM and NHK are also equal Then we have this place which is a four-way intersection where something like this happens That is EM and NHK are both parallel and equal Then that is the parallelogram EM and NHK are parallel and equal then that is the parallelogram Thus we obtain this figure is a parallelogram Now to prove that it's just a form Then people usually prove that there can be two types Prove that two adjacent sides are equal Two adjacent sides are equal You can also prove that the two diagonals are perpendicular Then it's easy to prove that the two diagonals are perpendicular here By doing this I will temporarily consider angle AMN This is an isosceles triangle Why is it an isosceles triangle Because I have AM and AN equal according to the proof given by teacher 1 Therefore this is an isosceles triangle What is H and what is MNH H is the midpoint the problem states that H is the midpoint Then for a median to the midpoint of the opposite side It is called a median When the triangle is equilateral the median is also the altitude The median is not only the altitude but it is also the perpendicular bisector and angle bisector combined Four lines are overlapping the median the altitude the perpendicular bisector overlapping so immediately we have AH The line is high and therefore the angle H is right This means that we will have the EK EK is perpendicular to MN A quadrilateral with two perpendicular diagonals is a rhombus Therefore I conclude that this is a rhombus Of course if I say it like this I'm not talking about right angles I'm saying that triangle AMN isosceles at A with H being the midpoint of MN Therefore AH is the median Therefore AH is also the perpendicular bisector Then I think AK is also the perpendicular bisector And therefore KM equals KN A quadrilateral with two adjacent sides equal is a kite Therefore we have proven statement 2 Theo the question of proving the perimeter A constant illusion Then we usually prove the sum of 3 sides Sum of 3 sides By something that does not change Then repeat here M is mobile Movement along edge BC Then it will also move in this area But this CMK illusion is an illusion where M and K move together But the idea is that the total of the three sides does not change Then we observe something very pleasant That is KM by KN KM equals KN That is why the three sides MC plus CK plus MK Will equal MC plus CK Now we replace KM with KN So does that mean the perimeter will become MC plus CK plus KN or does it mean it's just MC plus NC Right MC and NC are always together But this NC contains This DC It contains you DC and this DC is the side of the square that does not change So we will think that This MC and this ND when combined are great aren't they But we found out something which is also very nice This is the ND it is the same as the MB ND is equal to MB because these are two corresponding sides of equal simple shapes As I proved in the first statement I can immediately conclude that this is the case ND plus MC Then it is equal to BM plus MC That is equal to BC That is equal to BC and therefore it is the side of the square So in conclusion The perimeter of the MCK It will be equal to CD plus BC Therefore it is twice the side of the square Therefore it does not change And that is what we have proven See you again