M. What is the relationship between the area of Mn parallel to AC, △ ABM and △ BCN?

M. What is the relationship between the area of Mn parallel to AC, △ ABM and △ BCN?

The areas of △ ABM and △ BCN are equal

As shown in the figure, in the known parallelogram ABCD, e and F are the midpoint of AB and CD respectively, and points m and N are on AD and BC respectively, and am = CN
I have known that connecting AC, proving △ AEO ≌ △ COF; △ AOM ≌ △ CNO. (AAS)

ABCD is a parallelogram, ab = BC, ab = DC, a = C, B = D ∵ e, f is ab respectively. The midpoint of CD is AE = CF = be = DF ∵ am = cn ∵ DM = BN. In △ ame and △ CNF, a = CAE = CF, am = cn ≌ ame ≌ CNF ≌ EM = FN

As shown in the figure, square ABCD, M is a point on BC. Connect am, make the vertical bisector GH of am, intersect AB at point G, intersect CD at point h. given am = 10cm, find the length of GH

If the line segment GH is translated down to BN, then BN = GH, BN ⊥ am, ∵ quadrilateral ABCD is a square, ≁ AB = BC, ≁ ABC = ≁ BCD = ≁ BEM = 90 °. ≁ BAM = 90 ° - ≁ AMB = ≁ CBN. ≁ RT △ ABM ≌ RT △ BCN. Therefore, the center of square ABCD can be taken as the rotation center and rotated 90 ° counterclockwise to make RT △ ABM coincide with RT △ BCN, ≌ GH = BN = am = 10cm

As shown in the figure, square ABCD, M is a point on BC. Connect am, make the vertical bisector GH of am, intersect AB at point G, intersect CD at point h. given am = 10cm, find the length of GH

If the line segment GH is translated down to BN, then BN = GH, BN ⊥ am, ∵ quadrilateral ABCD is a square, ∥ AB = BC, ∥ ABC = ∥ BCD = ∥ BEM = 90 °. ∥ BAM = 90 ° - ∥ AMB = ∥ CBN. ∥ RT △ ABM ≌ RT △ BCN

As shown in the figure, square ABCD, M is a point on BC. Connect am, make the vertical bisector GH of am, intersect AB at point G, intersect CD at point h. given am = 10cm, find the length of GH

If the line segment GH is translated down to BN, then BN = GH, BN ⊥ am, ∵ quadrilateral ABCD is a square, ≁ AB = BC, ≁ ABC = ≁ BCD = ≁ BEM = 90 °. ≁ BAM = 90 ° - ≁ AMB = ≁ CBN. ≁ RT △ ABM ≌ RT △ BCN. Therefore, the center of square ABCD can be taken as the rotation center and rotated 90 ° counterclockwise to make RT △ ABM coincide with RT △ BCN, ≌ GH = BN = am = 10cm

In rectangular ABCD, ab = 3, BC = 5, fold △ CBE so that B falls on F of AD, CE is crease, and calculate the length of AF and be

Triangle CFD is easy to find that the three sides are 3,4,5, so DF = 4, AF = 1, and the rest be = EF. We can use Pythagorean theorem to get be = 5 / 3 in triangle AEF

ABCD is a giant, AB is equal to 4, ad is equal to 3, fold the rectangle along the straight line AC, point B falls on point E, connect De to find the area and perimeter of quadrilateral aced

Let the intersection of AE and CD be f
It is easy to get AC = 5
From folding, AE = AB = 4, EC = BC = 3
FA = FC is obtained from ∠ FCA = ∠ cab = ∠ fac,
Let FC = x, then EF = 4-x, in △ CEF, (4-x) ^ 2 + 9 = x ^ 2, solution x = 25 / 8, FC = 25 / 8, DF = 7 / 8
The easy proof of de ‖ AC is de: AC = DF: FC = 7:25, de = 7 / 5, perimeter is 12.4
The quadrilateral aced is an isosceles trapezoid, and its height is a right triangle. The height on the hypotenuse AC of ACE is 12 / 5
The area is (7 / 5 + 5) × (12 / 5) △ 2 = 192 / 25

ABCD is a rectangular line, ab = 4, ad = 3. Fold the rectangle along the straight line AC, and point B falls at point E, connecting de. the area? Perimeter? Of quadrilateral aced?

Area = 2.4 * (5 + 1.4) / 2 = 7.68 perimeter = 3 + 3 + 5 + 1.4 = 12.4

As shown in the figure, in rectangular ABCD, the angle DAE = angle CBE = 45 ° ad = 1, calculate the area and perimeter of △ Abe (accurate to 0.01)

4.83
AE=1.414.
BE=1.414
AB=2

As shown in the figure, in rectangular ABCD, AE ⊥ BD is in E, ad = 4cm, ∠ DAE = 2 ∠ BAE, then ∠ DAE=----------
（-√5）^2=
A (3, - 4) is in quadrant - ---, and the axisymmetric coordinates of X are--------------

∠DAE=60°
（-√5)^2=5
In the fourth quadrant, (3,4)