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As shown in the figure, in △ ABC, D is a point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
It is proved that in △ ACD, because ad = AC and AE ⊥ CD,
Therefore, according to the point of intersection between the vertical line of the bottom edge and the bottom edge of the isosceles triangle, namely the midpoint, it can be proved that:
E is the midpoint of CD, and because f is the midpoint of CB,
Therefore, EF ‖ BD, and EF is the median line of △ BCD,
So EF = 1
2bd, that is, BD = 2ef
In the triangle ABC, where the angle c is equal to twice, the angle B. D is a point on BC, and ad is perpendicular to ab. the point E is the midpoint of BD It is proved that the angle AEC is equal to the angle C It is proved that BD is equal to 2Ac 3 if AE is equal to 6.5 and ad is equal to 5, what is the circumference of the triangle Abe?
It is very simple. 1. Prove: angle c = 2 angle B, and angle AEC = angle B + angle EAB (exterior angle theorem), it can be seen that only angle B = angle EAB. Obviously, in the right triangle abd, according to the theorem, AE = be, then angle B = angle EAB. 2. According to the above angle c = angle AEB, AC = AE, so BD = 2ae = 2ac3
In the triangle ABC, angle B = angle c, D on BC, angle bad = 50 degrees, AE = ad, calculate the degree of angle EDC
∵AE=AD
∴∠ADE=∠AED
∵∠ADE+∠EDC=∠B+∠BAD=∠B+50°
∠B=∠C
∴∠AED+∠EDC=∠C+50
∵∠AED=∠EDC+∠C
∴∠EDC+∠C+∠EDC=∠C+50°
∴∠EDC=25°
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE (1) Verification: △ ABC ≌ △ ead; (2) If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED
(1) It is proved that: ∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC，AD=BC．
∴∠DAE=∠AEB．
∵AB=AE，
∴∠AEB=∠B．
∴∠B=∠DAE．
∵ in △ ABC and △ AED,
AB＝AE
∠B＝∠DAE
AD＝BC ，
∴△ABC≌△EAD．
(2) ∵ AE bisection ∵ DAB (known),
∴∠DAE=∠BAE；
∵ DAE= ∵ AEB,
∴∠BAE=∠AEB=∠B．
The △ Abe is an equilateral triangle
∴∠BAE=60°．
∵∠EAC=25°，
∴∠BAC=85°．
∵△ABC≌△EAD，
∴∠AED=∠BAC=85°．
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE. If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED Verification: △ ABC ≌ △ EAD
Proof: in the parallelogram ABCD
AD∥BC AD=BC
∴∠DAE=∠AEB
∵∠BAE=∠DAE
∴∠BAE=∠AEB
∵AB=AE
∴∠ABE=∠AEB
∴∠ABE=∠AEB=∠BAE
ν Δ Abe is an equilateral triangle
∴∠ABE=∠AEB=∠BAE=∠DAE=60°
AB=BE=AE
In △ ABC and △ AED
AB=AE
∠ABE=∠DAE
BC=AD
∴△ABC≌△AED (SAS)
∴∠AED=∠BAC=∠BAE+∠EAC
=60°+25°
=85°
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE (1) Verification: △ ABC ≌ △ ead; (2) If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED
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According to the meaning of the title, Abe is a regular triangle
In addition, if ∠ ADC = ∠ B = ∠ DAE, then adce is isosceles trapezoid,
Then, BAE = - 85 ° + BAC = 25 °
As shown in the figure, in the parallelogram ABCD, e is a point on the edge of BC, and ab = AE (1) Verification: △ ABC ≌ △ ead; (2) If AE bisects ∠ DAB, ∠ EAC = 25 °, find the degree of ∠ AED
(1) It is proved that:
As shown in the figure, in △ ABC, D is the midpoint of BC, the bisector AE of the intersection angle BAC of de ⊥ BC is at e, EF ⊥ AB is at F, and the extension of AC intersected by eg ⊥ AC is proved to be BF = CG Can I extend BC and eg for urgent use? I feel I'm right
It is unnecessary for AE to be an angular bisector, EF ⊥ AB, eg ⊥ AC, the angle AEF = angle AEG
Because the angle bed = angle CED, the angle bef = angle CEG, be = EC, angle BFE = angle CGE
So the triangle bef is congruent with the triangle CEG, so BF = CG
It's too much of a stretch
In the triangle ABC, D is the midpoint of BC, ED is perpendicular to BC, the angle bisector of BAC is perpendicular to point e EF is perpendicular to point F eg is perpendicular to AC, and the extension of AC is at point G In this way, we can prove that if we join the two right angles on the lower edge of the graph AF, we can find that if we join the two right angles on the Acb, we can find that the two edges are connected together,
Connecting EC, EB
Because EA is the bisector of the angle cab
It is known that EF is perpendicular to point F, eg is perpendicular to AC, and the extension of AC is at point G
Therefore, it is easy to know that eg = EF
There is also ed vertical bisection BC
It is also easy to know that EC = EB
So two right triangles CGE and BFE are congruent
So BF = CG
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
- 2. As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad
- 3. In the triangle ABC, the points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and the ad bisection angle BAE (1) when the angle BAC is equal to 90 degrees In the triangle ABC, points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and ad bisects the angle BAE (1) When the angle BAC is equal to 90 degrees, it is proved that BD is equal to AC (2) When the angle BAC is not equal to 90 degrees, is there still BD equal to AC? Explain the reason
- 4. As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De
- 5. 0
- 6. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 7. It is known that AC and BD are two mutually perpendicular chords of the circle O: x2 + y2 = 4, and the perpendicular foot is m (1, 2) The maximum area of the quadrilateral ABCD is () A. 4 B. 4 Two C. 5 D. 5 Two
- 8. It is known that AB is the diameter of ⊙ O. the straight line BC and ⊙ o are tangent to point B. through a, ad ⊙ OC intersects ⊙ o at point D and connects CD Proof: CD is tangent of ⊙ o
- 9. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
- 10. As shown in the figure, it is known that AB is parallel to CD, OA = OD, BC passes through point O, e, f are on the straight line AOD and AE = DF
- 11. As shown in the figure, in diamond ABCD, AE ⊥ BC, perpendicular foot is e, angle DAE = 2, angle BAE, BD = 15, find the length of AC, ab
- 12. As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle
- 13. As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC
- 14. In △ ABC, ∠ a = 1 2∠B=1 3 ∠ ACB, CD is the height of △ ABC, CE is the angular bisector of ∠ ACB, and the degree of ∠ DCE is calculated
- 15. It is known that the triangle ABC, CA = CB, point O is on the vertical bisector of Ca and CB, and m.n is on a straight line AC.BC Go ahead, It is proved that CN + Mn = am
- 16. As shown in the figure, in the RT triangle ABC, the angle c = 90 degrees, the DB bisector angle ABC intersects AC at point D, De is the vertical bisector of AB and intersects AB at point E (1) Find the degree of angle A (2) If BC = 6, AC = 8, then the circumference of the triangle BDC is
- 17. 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
- 18. As shown in the figure, ∠ ABC = ∠ DBE = 90 °, DB = be, ab = BC (1) Confirmation: ad = CE, ad ⊥ CE; (2) If the external conditions of △ DBE rotating around point B to △ ABC remain unchanged, then the conclusion in (1) still holds? Draw a graph to prove your conclusion
- 19. In the triangle ABC, BD = DC = AC, D is the midpoint of BC and E is the midpoint of DC. It is proved that the bisection angle BAE of AD
- 20. 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