As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm, now fold △ ABC to make vertex A and B coincide, then the length of crease De is () cm A. 52B. 154C. 158D. 5

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm, now fold △ ABC to make vertex A and B coincide, then the length of crease De is () cm A. 52B. 154C. 158D. 5

Let AE = x, be = x, CE = 4-x, be = x, CE = 4-x, be = x, CE = 4-x, be = x, CE = 4-x, be = 4-x, be = x, CE = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 2, be = 4-x, be = 2, be = 2, be = 4-x = 258, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 2, be =

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm, now fold △ ABC to make vertex A and B coincide, then the length of crease De is () cm
A. 52B. 154C. 158D. 5

Let AE = x, be = x, CE = 4-x, be = x, CE = 4-x, be = x, CE = 4-x, be = x, CE = 4-x, be = 4-x, be = x, CE = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 2, be = 4-x, be = 2, be = 2, be = 4-x = 258, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 4-x, be = 2, be =

As shown in the figure, it is known that in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm. Now fold △ ABC to make vertex A and B coincide, and find the length of DB and crease De

Let d be the intersection point of AC and E be the intersection point of ab. according to the graph, let ad = BD be the intersection point of AB and de = 25 / 8 be the intersection point of ab. according to the Pythagorean theorem, let de be the intersection point of AC and ab. because in rtabc, AC = 4 and BC = 3, according to the Pythagorean theorem, let AB = 5 be the vertical bisector of AB because a and B should coincide, So ad = BD let CD be x, then there is: 4-x = the square of X under the root sign plus the square of 3, that is: 4-x = the square of X under the root sign + 9. Solve the equation: the square of (4-x) under the root sign = the square of X + 9 (remove the root sign) to get: (4-x) the square = the square of X + 9, then: 16-8x + x the square = the square of X + 9, 16-8x = 9, 8x = 7, 8x = 7 / 8, so ad = AC - = 4-7 / 8 = 25 / 8 in the right triangle ade, AE = 5 / 2 (you know the nature of the vertical bisector), ad = 25 / 8 (just obtained) According to the Pythagorean theorem, the square of ad minus the square of AE under de = root is obtained: de = 15 / 8

A piece of right triangle paper ABC, ∠ A is equal to 90 °, AC = 3, ab = 4, folded so that points B and C coincide, and the crease intersects with AB and BC at points D and e respectively

The crease intersects with AB and BC at points D, e respectively,
Point E is the midpoint of BC
If the crease and ab intersect at m, then me is parallel to AC, which is the median line of triangle ABC, me = 1.5
At this point, the triangle MPa is a right triangle, and the two right sides MP = 1.5, am = 4 divided by 2 = 2, and you want to find the hypotenuse AE
AE=2.5

M is a point on edge AB in triangle ABC, and am ^ 2 + BM ^ 2 + cm ^ 2 = 2am + 2bm + 2cm-3, then AC ^ 2 + BC ^ 2 =?

Am ^ 2 + BM ^ 2 + cm ^ 2 = 2am + 2bm + 2cm-3 am ^ 2 + BM ^ 2 + cm ^ 2-2am + 2bm + 2cm + 3 = 0 (am ^ 2 + 2am + 1) + (BM ^ 2 + 2bm + 1) + (cm ^ 2 + 2cm + 1) = 0 (am-1) ^ 2 + (BM-1) ^ 2 + (cm-1) ^ 2 = 0 am-1 = 0 BM-1 = 0 cm-1 = 0 am = 1 BM = 1 cm = 1 AC ^ 2 + BC ^ 2 = AB ^ 2 = (am + BM) ^ 2 =

As shown in the figure, △ ABC is an equilateral triangle, △ DBC is an isosceles triangle with a vertex angle of 120 ° and D as the vertex to make an angle of 60 ° with two sides of the angle respectively
B. AC connects Mn at two points m and N, and proves that Mn = BM + CN

As shown in the figure, take MD as the axis, reverse △ BMD to get △ EMD, then BM = me, BD = de = CE
When the me cross AC to n ·, then
RT△EDN·≌RTCDN·,
So Mn · = BM + & nbsp; CN·
It is not difficult to prove that ∠ 2 + ∠ 3 = 60 ° so n point coincides with n

Triangle ABC is an equilateral triangle, extending BC to D and E in both directions, making the angle DAE = 120 ° if DB = 9 and CE = 4, find s △ ade

If D is on the left side of B and E is on the right side of C, ∠ DAB = ∠ 1, ∠ EAC = ∠ 2, let the side length be x ∵ equilateral triangle ABC, ≠ BAC = ∠ ABC = ACB = 60 ∵ 1 + ∠ 2 = ∠ 1 + ∠ d = ∠ 2 + ∠ e = 60 ∵ 1 = ∠ e, ∠ 2 = ∠ D ∵ triangle DBA ∵ triangle ace ∵ column proportion formula is x = 6 ∵ de =

It is known that the triangle ABC is an equilateral triangle, the points D, B, C and E are on the same line, and the angle DAE is equal to 120 degrees

Δ ABC is an equilateral triangle, so ∠ BAC = ∠ ABC = ∠ ACB = 60
Because ∠ DAE = 120, ∠ DAB + ∠ EAC = 60
If ∠ ABC is the outer angle of △ DAB, then ∠ DAB + ∠ ADB = ∠ ABC = 60
So ∠ ADB = ∠ EAC
∠DBA=∠ACE=120
So △ abd ∽ ECA
BD：AC=AB：CE,BD×CE=AC×AB
Because AB = AC = BC, BD × CE = BC & # 178;

As shown in the figure, △ ABC is an equilateral triangle, ∠ DAE = 120 ° to prove: (1) △ abd ∽ ECA; (2) BC2 = DB · CE

It is proved that: (1) the ∵ ABC is an equilateral triangle, ∵ DAE = 120 °, the ∵ DAB + CAE = 60 °, the ∵ ABC is the outer angle of ∵ abd, ∵ DAB + D = ∵ ABC = 60 °, the ∵ CAE = ∵ D, ∵ ABC = ∵ ACB = 60 °, the ∵ abd = ∵ ace = 120 °, the ∵ abd ∽ ECA; (2) the ∵ abd ∽ ECA

As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector D of AB intersects AC at e, the vertical bisector of CE just passes through point B, intersects a at point F, and calculates the degree of angle a?

Connect be
Because De is the vertical bisector of AB, angle a = angle DBE
Since BF is the vertical bisector of EC, angle c = angle bec
Angle a + angle c + angle ABC = 180 = > angle a + 2 angle c = 180 (1)
Angle EBC = 180-angle bec-angle C = 180-2 angle c (2)
Angle ABC = angle Abe + angle EBC = > angle c = angle a + angle EBC (3)
From (2) (3), angle c = angle a + 180-2, angle c (4)
According to (1) (4), the angle c is 72 degrees and the angle a is 36 degrees