As shown in the figure, in the triangle ABC, < a = 55 ° and H is the intersection of high BD and EC, then < BHC=
<BHC=<DHE=360-90-90-55=125
RELATED INFORMATIONS
- 1. It is known that, as shown in the figure, in △ ABC, ∠ ABC = 66 ° and ∠ ACB = 54 ° be and CF are the heights of AC and AB on both sides, and they intersect at point h. calculate the degree of ∠ Abe and ∠ BHC
- 2. As shown in the figure, in the triangle ABC, angle A: angle ABC: angle ACB = 4:5:6, BD and CE are the intersection points h of height, BD and Ce on AC and ab respectively, and the degree of angle BHC is calculated
- 3. As shown in the figure, in the triangle ABC, the known angle ABC is 66 degrees, the angle ACB is 54 degrees, be is the height of AC, h is the intersection of be and CF, and the angle Abe and angle ACF are calculated Degree of sum angle bec
- 4. As shown in the figure, D is a point on the edge BC of △ ABC, ab = 2, BD = 1, DC = 3. Prove: △ abd ∽ CBA
- 5. D is a point on the bisector of an outer angle of the triangle ABC, connecting dB and DC to prove AB + AC < DB + DC
- 6. In the triangle ABC, ad is the angle bisector, ab = 5, AC = 4, BC = 7. Please use the method of similar triangle to find the length of BD and DC
- 7. In Δ ABC, ∠ C 〉 ∠ B, ad is perpendicular to point D, BC is equal to point D, AE is equal to ∠ BAC, please specify ∠ DAE = 1 / 2 (∠ B ∠ C)
- 8. As shown in the figure, in △ ABC, ∠ C = 90 °, the vertical bisector of AB intersects at point D, AB intersects at point E, the degree ratio of ∠ DAE and ∠ DAC is 2:1, and the degree of ∠ B is calculated
- 9. As shown in the figure, in △ ABC, ∠ C = 90 °, the vertical bisector of AB intersects at point D, AB intersects at point E, the degree ratio of ∠ DAE and ∠ DAC is 2:1, and the degree of ∠ B is calculated
- 10. In △ ABC, be bisects ∠ ABC, ad is the height on BC, and ∠ ABC = 60 ° and ∠ BEC = 75 ° to calculate the degree of ∠ DAC
- 11. As shown in the figure, in △ ABC, points E and G are on BC and AC respectively, CD ⊥ AB and ef ⊥ AB, and the perpendicular feet are D and f respectively (1) Is CD parallel to ef? Why? (2) The degree of ∠ ACB can be calculated when ∠ 1 = 2, ∠ 3 = 105 ° is known
- 12. It is known that in △ ABC, e and F are the midpoint of AB and AC respectively, CD bisects ∠ BCA and intersects EF with D Verification: ad ⊥ DC Why AF = CF = EF, ADC = 90
- 13. In the parallelogram ABCD, AE bisection ∠ DAB intersects DC with E, BF bisection ∠ ABC intersects DC with F, DC = 8cm, ad = 3cm, find the length of De, DF and FC RT
- 14. If ∠ BDC = x + 2 / 3 ∠ a, find the degree of X?
- 15. Known: as shown in the figure, ∠ abd = ∠ DBC, ∠ ACD = ∠ DCE. (1) if ∠ a = 50 °, find the degree of ∠ D; (2) guess the relationship between ∠ D and ∠ a, and explain the reason; (3) if CD ‖ AB, judge the relationship between ∠ ABC and ∠ a
- 16. If ∠ BDC = α + 2 / 3 ∠ a, α is obtained
- 17. In △ ABC, Sina * SINB = cos ^ 2 (C / 2), C = 4, C = 40 ° to find the value of A
- 18. As shown in the figure, it is known that AB / / CD, CE divide ∠ ACD equally, hand AB to e, ∠ a = 118 °, please fill in the reason for ∠ 2 = 31 °
- 19. Known: as shown in the figure, CE bisects ∠ ACD, ∠ 1 = ∠ B, verification: ab ‖ CE
- 20. As shown in figure a, CE / / AB, so ∠ 1 = 2 = B, so ∠ ACD = 1 + ∠ 2 = a + B, which is a useful conclusion, In the quadrilateral ABCD of figure B, do AE / / BC to intersect DC with e through A. if you have this conclusion, find the degree of ∠ a + ∠ B + ∠ C + ∠ D Figure a http://hiphotos.baidu.com/yalijudy/pic/item/099efed1bbb16a299a50274f.jpg Figure B http://hiphotos.baidu.com/yalijudy/pic/item/6750e5f54b044b34bd3109e2.jpg