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As shown in the figure, in the isosceles triangle ABC, ab = AC, D is a point on the extension line of AB, BD = AB, CE is the center line on waist AB, and verify; CD = 2ce needs specific process
Make the center line BF of Δ ABC,
∵AB=AC,AE=1/2AB,AF=1/2AC,
ν AE = AF, and ∠ a = ∠ a,
∴ΔABF≌ΔACE,∴CE=BF,
∵ BF is the midpoint of AD and WC respectively,
ν BF is the median line of Δ ADC,
∴CD=2BF,
∴CD=2CE.
It is known that in △ ABC, ab = AC, BD and CE are the heights on the edge of AC and ab respectively, connecting De Results: (1) △ abd ≌ △ ace; (2) The quadrilateral BCDE is isosceles trapezoid
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It is proved that: (1) ∵ BD and CE are the heights of AC and ab respectively
And ∵ a = ∠ a, ab = AC,
∴△ABD≌△ACE；
(2) If ad = AE is obtained from △ abd ≌ △ ace, then ∠ ade = ∠ AED,
Therefore, ∠ ade = 180 ° − a
2．
∵ AB = AC, we get ∵ ABC = ∠ ACB, so ∠ ACB = 180 ° − a
2．
∴∠ADE=∠ACB．
∴DE∥BC．
And ∵ ab-ae = ac-ad, that is be = CD,
The quadrilateral BCDE is isosceles trapezoid
In the isosceles triangle ABC, ab = AC, BD ⊥ AC, CE ⊥ AB, the vertical feet are points D, e, and link de. it is proved that the quadrilateral BCDE is isosceles trapezoid—— On the first floor, the congruence of RT triangle can be directly used HL, and the angle of the last step has nothing to do with parallelism.
Parallel process:
∵EB=DC,∴AE=AD,∴∠AED=∠ADE=∠ABC=∠ACB
∴ED‖BC
As shown in the figure, in the isosceles triangle ABC, ad is equal to AC BD, vertical AC CE, and vertical AB feet are points D. e. the quadrilateral BCDE is isosceles trapezoid
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It is proved that: (1) BD and CE are the heights of AC and ab respectively, and ∵ a = a, ab = AC, abd ≌ △ ace; (2) if ad = AE is obtained from △ abd ≌ △ ace, then ∵ ade = 180 ° − A2. ∵ AB = AC, so ? ACB = 180 ° − A2
It is known that in the isosceles triangle ABC, CE, ab = AC, D, e are points on AC and ab respectively and be = CD, it is proved that BD = CE It is known that in the isosceles triangle abcce, ab = ACDE are the points on acab respectively and be = CD. It is proved that BD = CE
Because AB = AC, so the angle ABC = angle ACB, and because be = CD, BC = BC, so the triangle BCE is all equal to CBD, so BD = CE
In this paper, we prove that the vertex of △ on the intersection line of ABC is an isometric point
It is proved that: the over point D is the DG
It is known that in △ ABC, ∠ B = 90 °, ab = BC, D is on AB, e is on BC, BD = CE, M is the midpoint of AC, it is proved that △ DEM is isosceles right triangle
If BM is connected, then BM=MC, ﹤ DBM= ﹤ C=45 °
And BD = CE = = = > △ BDM ≌ △ CEM = = > MD = me
The △ DEM is an isosceles right triangle
As shown in the figure, in the isosceles right triangle ABC, ∠ C = 90 ° AC = BC, points D and E are on BC and AC respectively, and BD = CE, M is the midpoint of ab Is △ MDE isosceles RT △ then?
Because m is ab, midpoint MC = MB, so MEC of triangle is congruent triangle MDB (SAS), me = MD, angle EMC = angle DMB, so angle CMD + angle DMB = angle EMC + angle CMD = 90 degrees, so EMD is isosceles triangle
RELATED INFORMATIONS
- 1. In △ ABC, ab = BC, ∠ B = 90 °, M is the midpoint of AC, D.E is the moving point on AB and BC respectively, and BD = CE, the shape of △ DEM is obtained and proved
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