As shown in the figure, △ ABC has rhombic ampn, if ammb = 12, then BPBC=______ ．

As shown in the figure, △ ABC has rhombic ampn, if ammb = 12, then BPBC=______ ．

Therefore, the answer is: 23

It is known that the intersection of diagonal lines AC and BD in quadrilateral ABCD at O m.n is OA.OC Midpoint proof BM = DN BM parallel dn

Isn't that right? If you draw a trapezoid or other irregular quadrilateral, it's definitely wrong. Let's prove the parallelogram? The diagonals of parallelogram are equally divided. You know, so om = on, OB = od and opposite vertex angle, so they are congruent, so BM = DN, angle MBO = angle ndo, so they are parallel

As shown in the figure, the points m and N are on the sides BC and CD of the square ABCD respectively. It is known that the perimeter of △ MCN is equal to half of the perimeter of the square ABCD, then ∠ man=______ ．

When △ adn is rotated 90 ° clockwise around point a, we can get △ Abe,  AE = an, be = DN,  Abe =  d = 90 degree,  NAE = 90 degree and  ABC = 90 degree,  points m, B and E are collinear,  me = be + BM = DN + BM,  MCN's perimeter is equal to half of square ABCD's perimeter,  Mn + NC + MC = DC + BC = DN + NC + MC + BM,  Mn = DN + BM,  Mn = me. In △ man and △ Mae, an = aemn = meam = am,  Mn = me Therefore, the answer is 45 degrees

As shown in the figure, points m and N are on the sides BC and CD of the square ABCD respectively. Given that the perimeter of the triangle MCN is equal to half of the perimeter of the square ABCD, the degree of the angle man is calculated
The detailed process of proving the angular equality of the second congruence

When △ adn is rotated 90 ° clockwise around point a, we can get △ Abe,  AE = an, be = DN,  Abe =  d = 90 degree,  NAE = 90 degree, and  ABC = 90 degree,  points m, B and E are collinear,  me = be + BM = DN + BM,  △ MCN is equal to half of the circumference of square ABCD,  Mn + NC + MC = DC + BC = DN + NC + MC + BM

The points Mn are on the sides BC and CD of the square ABCD respectively. Given that the perimeter of the triangle MCN is equal to half of the perimeter of the square ABCD, the degree of ∠ man is calculated?
Which theorem of square can be proved AN.AM Is the angular bisector of ∠ DAE and ∠ DAB,

From the question ~ we can get the relation ~ Mn = DM + BM, take a point E on Mn, so that DM = em, BN = en. Because it is a square, then we can get that am and an are the angular bisectors of angle DAE and DAB respectively ~ so the angle man = ∠ dam + ∠ ban, so the angle man is 45 degrees~

In the parallelogram ABCD, am ⊥ BC, an ⊥ CD, m and N are perpendicular feet. If AB = 13, BM = 5 and MC = 9, the length of Mn is______ ．

In ∵ am ⊥ BC, ab = 13, BM = 5, ∵ RT △ ABM, am = AB2 − BM2 = 12, SINB = amab = 1213, ∵ quadrilateral ABCD is parallelogram, ∵ d = ∵ B, ∵ an ⊥ CD, ad = BC = BM + MC = 14, ∵ in RT ⊥ adn, sind = anad = 1213, ∵ an = 16813, ∵ amn, am = 12, ∵ B + ∠ ba

M. What is the relationship between the area of △ adn and the area of △ ABM

Two triangles have the same area,
Method 1
Suppose that m and N are the midpoint of BC and CD, and meet the condition of Mn / / BD; let ad side length be a, BC side length be B; the height of vertical ad side is H1, and the height of vertical BC side is H2, then the area of parallelogram is s = a * H1 or S = b * H2; that is, there is a * H1 = b * H2; because m and N are the midpoint respectively, then there is NC = 0.5cd = 0.5bc = 0.5B; cm = 0.5ad = 0.5A; thus the area of quadrilateral ABCN is S1 = 0.75b * H2; the area of quadrilateral AMCD is S2 = 0.75a * H1; the area of quadrilateral ABCN is S1 = 0.75b * H2; the area of quadrilateral AMCD is S2 = 0.75a * H1; That is, S1 = S2; and because the area of triangle ABM S3 = s-s2;
The area of triangle adn is S4 = s-S1; so S3 = S4; that is to say, the area of triangle ABM and triangle adn are equal
Method 2
Suppose that m and N are infinitely close to B and D respectively; that is, Mn is infinitely close to BD, then the areas of triangle adn and triangle ABM are equal, close to 0; (or suppose that two points are infinitely close to C, then the areas of two triangles are close to half of parallelogram, that is, equal)

M. N is the side BC of parallelogram ABCD and the point on CD, Mn is parallel BD, and the area size relationship of triangle adn and triangle ABM

Two triangles are equal in area. Method 1: suppose that m and N are the midpoint of BC and CD, which meets the condition of Mn / / BD; let ad side length be a, BC side length be B; the height of vertical ad side is H1, and the height of vertical BC side is H2, then the area of parallelogram is s = a * H1 or S = b * H2; that is, there is a * H1 = b * H2; because m and N are the midpoint respectively, then

As shown in the figure, m and N are the points on the edge AD and CD of the parallelogram ABCD respectively, and Mn ‖ AC, then what is the relationship between the areas of △ ABM and △ BCN? Try to explain the reason

If ∥ m and N are the points of AD and CD respectively, and Mn ∥ AC, then AMAD = cncd, the area of ∥ ABM is 12 AMAD times that of parallelogram ABCD, the area of ∥ BCN is 12 cncd times that of parallelogram ABCD, the area of ∥ AMAD = cncd, the area of ∥ ABM and ∥ BCN are equal

In the parallelogram ABCD, Mn parallel BD intersects DC in N and BC in M. the area relationship between triangle and triangle ABM is compared to explain the reason

Equal relation. As long as another diagonal AC can explain the problem