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This is the geometry of grade two, which is related to parallelogram In the RT triangle ABC, the angle c = 90 degrees, M is the midpoint of AB, am = an, Mn is parallel to AC, and Mn = AC
Connect cm, because cm is the center line on the hypotenuse of a right triangle, so cm = am, so, ∠ MAC = ACM
And am = an, so amn = anm
Mn / / AC, so ∠ MAC = amn
Therefore, AMC = man
So, an / / cm
Therefore, the quadrilateral acmn is a parallelogram, so Mn = AC
As shown in the figure, in RT △ ABC, ∠ C = 90 ° m is the midpoint of AB, am = an, Mn ‖ AC, is it true to guess Mn = AC? Why?
It is established;
If cm is connected, CM = am (the center line of the hypotenuse of a right triangle is equal to half of the hypotenuse);
And because AC ∥ Mn, the angle amn = cam;
Because am = an = cm; that is, amn and AMC are isosceles triangles, and the base angles of the two triangles are equal and the waist is equal, then the two triangles are congruent;
AC = Mn
In the RT triangle ABC, the angle c = 90 degrees, M is the midpoint of AB, am = an, Mn is parallel to AC, if the condition am = an is changed to am perpendicular to an
Add condition AC = BC
prove:
∵ ACB = 90 ° and M is the midpoint of ab
∴CM⊥AB
∵AN⊥AB
∴AN∥CM
∵AC∥MN
The quadrilateral acmn is a parallelogram
∴MN=AC
It is known that in RT △ ABC, ∠ C = 90 ° m is the midpoint of AB, am = an, Mn ‖ AC Mn = AC
Proof: as shown in the figure, connect cm, (1 point)
∵∠ACB=90°，
∴CM=AM=1
2AB，
﹤ MAC = ∠ MCA, (1 point)
∵ am = an, amn = ∠ n, (1 point)
∵MN∥AC，
∴∠NMA=∠MAC，∠CAN+∠N=180°，
∴∠CAN+∠MCA=180°，
∥ cm, (2 points)
The quadrilateral acmn is a parallelogram (1 point)
/ / Mn = AC. (1 point)
As shown in the figure, ∠ C = 90 ° in △ ABC is the vertical bisector of AB, and ∠ bad: ∠ CAD = 2:1, then ∠ B=______ .
In ∵ ABC, ∵ ACB = 90 ° and De is the vertical bisector of ab,
ν ad = BD, i.e. ∠ bad = ∠ abd,
∵∠BAD：∠CAD=2：1，
Let ∠ bad = x, then ∠ CAD = X
2，
∵ bad + CAD + ∠ abd = 90 °, i.e. x + X
2+x=90°，
The solution is: x = 36 °,
∴∠B=36°．
So the answer is 36 degrees
As shown in the figure, in △ ABC, ∠ C = 90 ° the vertical bisector of AB intersects BC at D, the perpendicular foot is e, ∠ CAD: ∠ DAB = 2:5 Find the degree of ∠ bac
No picture. You take your picture
52.5°
Let the degree of ∠ BAC be X
Then / 2x = 7
Since De is the vertical bisector of AB, DAB = ∠ DBA = 5x / 7
So we have 90 ° - x = 5x / 7
X = 52.5 ° is obtained
In △ ABC, the vertical bisector of ad ⊥ BC, BC intersects AC with E, be crosses ad with F. it is proved that e is on the vertical bisector of AF
It is proved that the vertical bisector of ∵ BC intersects AC and E,
∴BE=CE，
∴∠EBC=∠C，
∵AD⊥BC，
∴∠C+∠CAD=90°，∠EBC+∠BFD=90°，
∴∠CAD=∠BFD，
∵∠BFD=∠AFE，
∴∠AFE=∠CAD，
∴AE=EF，
﹤ e is on the vertical bisector of AF
In △ ABC, the vertical bisector of ad ⊥ BC, BC intersects AC with E, be crosses ad with F. it is proved that e is on the vertical bisector of AF
It is proved that the vertical bisector of ∵ BC intersects AC and E,
∴BE=CE，
∴∠EBC=∠C，
∵AD⊥BC，
∴∠C+∠CAD=90°，∠EBC+∠BFD=90°，
∴∠CAD=∠BFD，
∵∠BFD=∠AFE，
∴∠AFE=∠CAD，
∴AE=EF，
﹤ e is on the vertical bisector of AF
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, De is the vertical bisector of AB, and de intersects the extension lines of AB, AC and BC respectively at D, e, F, and cosa = 4 / 5, CE = 3
Therefore (4t-0) / [(6-3t) - (- 4 + 2t) / [(6-3t) - (- 4 + 2t) / [(6-3t) - (- 4 + 2t)] = 3 / 4, the equation is: T = 30 / 31, therefore, when t = 30 / 31, Mn is vertical ab. (x0d (2) MP / PN ratio will not change. The reasons are as follows: x0d pass through N point as nd vertical X axis, the intersection point is vertical X axis, the intersection point is vertical X axis, the intersection point is vertical X axis X axis, the intersection point is vertical X axis X axis, the intersection point is vertical X axis, the intersection point is vertical X axis X axis, the intersection point is vertical X axis, the intersection point is vertical X axis, the intersection point is vertical X axis X axis, the reason is as follows:the coordinate of point D is d (6-3t, 0), \In right triangle MND, MP / PN = Mo / OD, x0d and Mo = 4-2t, OD = 6-3t, (0
As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, BC = 3, AE = 4, the vertical bisector De of AB intersects the extension line of BC at point E, then what is the CE length?
What does this... Do for RT △ s
A point on a vertical bisector that is equidistant from both ends of the line segment
∴AE=BC
∵BC=3,AE=4
∴CE=1
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