![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector De of AB intersects AC at point E, and the vertical bisector of CE just passes through point B and intersects AC at point F. the degree of ∠ A is calculated
∵ △ ABC is an isosceles triangle, ∵ ABC = ∠ C = 180 − A2 ①, ∵ De is the vertical bisector of line AB, ∵ a = ∠ Abe, ∵ CE's vertical bisector just passes through point B and intersects with AC at the point, we can see that ∵ BCE is an isosceles triangle, ∵ BF is the bisector of ∠ EBC, ∵ 12 (∠ ABC - ∠ a) + ∠ C = 90 °, that is, 12 (∠ C - ∠ a) + ∠ C = 90 ° ②, ① ② is combined, so ∠ a = 3 6°．
RELATED INFORMATIONS
- 1. As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector De of AB intersects AC at point E, and the vertical bisector of CE just passes through point B and intersects AC at point F. the degree of ∠ A is calculated
- 2. As shown in the figure, in △ ABC, it is known that ∠ DBC = 60 ° AC ＞ BC, and △ ABC ′, △ BCA ′, △ cab ′ are equilateral triangles outside △ ABC shape, and point D is on AC, and BC = DC (1) prove: △ C ′ BD ≌ Δ B ′ DC; (2) prove: △ AC ′ D ≌ Δ DB ′ a
- 3. As shown in the figure, in the triangle ABC, angle a is equal to 60 degrees, angle B is equal to 75 degrees, ab = 4cm, find C of BC
- 4. As shown in the figure, ad is the middle line of △ ABC, point E is on the extension line of BC, CE = AB, ∠ BAC = ∠ BCA, verification: AE = 2ad
- 5. As shown in the figure, in the triangle ABC, ab = BC, ad is the middle line on BC, extend BC to point E, make CE = BC, prove: AE = 2ad
- 6. As shown in the figure, in the triangle ABC, the angle BAC = 90 ° and BD is the bisector of the angle ABC. The extension line of BD is perpendicular to the line passing through point C and E, and the extension line of CE intersecting Ba is F. the proof is: BD = 2ce Please help me
- 7. In RT △ ABC, ∠ C is equal to 90 degree CD.CE If ∠ A is equal to 30 °, try to judge the shape of ∠ BCE Geometric steps, is an equilateral triangle
- 8. In RT △ ABC, CD is the bisector of BCE. BCE = 60 °∠ 3 = ∠ 4. Describe the relationship between CB and CE. What's your conjecture? And prove it Note that it's in the RT triangle
- 9. As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, point D is a point on the hypotenuse AB, make ∠ CDE = ∠ a, pass through point C, make CE ⊥ CD, cross De to e, connect be Verification: ab ⊥ be
- 10. In △ ABC, ab = BC, ∠ a = ∠ C = 45 ° rotate △ ABC clockwise around point B by an angle α (0 ＜ α ＜ 90 °) and △ MBN, BM intersect AC with point E, Mn intersect AC, BC with D and f respectively (1) When α =? EM is the longest; (2) Observe and guess the quantitative relationship between EM and FC in the process of rotation, and explain the reason There should be an answer before 6:30 tomorrow, otherwise it will be invalid
- 11. As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector De of AB intersects AC at point E, the perpendicular foot is D, and the vertical bisector of CE is just right When passing through point B, the perpendicular foot is f, then the degree of ∠ A is—— I'm in a hurry. I still have a lot of homework. There are only a few questions left Help, I'm too lazy to do it. You should draw it out. The answer is 36 degrees. How can you figure out what it is like, give it to you, and hope you can give it awesome.
- 12. As shown in the figure, it is known that in △ ABC, ab = AC, D is a point on the CB extension line, ∠ ADB = 60 °, e is a point on ad, and de = dB
- 13. As shown in the figure, in △ ABC, ab = AC, points D and E are two points on the edge of BC, and ad = BD, AE = CE. If ∠ DAE = 60 °, calculate the degree of ∠ BAC
- 14. As shown in the figure, △ ABC is an equilateral triangle, ∠ DAE = 120 ° to prove: (1) △ abd ∽ ECA; (2) BC2 = DB · CE
- 15. As shown in the figure, point D is a point on the AC side of equilateral △ ABC, ∠ 1 = ∠ 2, BD = CE
- 16. As shown in the figure, point D is a point on the AC side of equilateral △ ABC, ∠ 1 = ∠ 2, BD = CE
- 17. Triangle ABC is an equilateral triangle, D, B, C, e are on a straight line, angle DAE = 120 degrees, known BD = 1, CE = 3. Find the side length of an equilateral triangle
- 18. As shown in the figure, △ ABC and △ DBC, the angle BAC = angle BDC = 90 °, O is the midpoint of BC (1) △ AOD is isosceles triangle? Why?
- 19. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 °, point m is on edge BC, am = BM
- 20. In RT △ ABC, ∠ B = 90 ° is connected to the midpoint m of AC side, and BM = cm = am is proved