As shown in the figure, D is a point on the edge BC of △ ABC, ab = 2, BD = 1, DC = 3. Prove: △ abd ∽ CBA

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• 1. 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
• 2. 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
• 3. 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)
• 4. 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
• 5. 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
• 6. In △ ABC, be bisects ∠ ABC, ad is the height on BC, and ∠ ABC = 60 ° and ∠ BEC = 75 ° to calculate the degree of ∠ DAC
• 7. It is known as follows: as shown in the figure, in △ ABC, ad is the height on the edge of BC, AE is the bisector of ∠ BAC, ∠ B = 50 °, DAE = 10 °, (1) find the degree of ∠ BAE; (2) find the degree of ∠ C
• 8. As shown in the figure, in △ ABC, ad ⊥ BC is at point D, AE bisects ⊥ BAC intersects BC at e, ⊥ C = 70 ° and ⊥ B = 38 ° to find the degree of ⊥ DAE
• 9. 1. As shown in the figure: in the RT triangle ABC, the angle ABC = 90, Ba = BC. Point D is the midpoint of AB, CD, through point B as BC, vertical CD 1. As shown in the figure: in the RT triangle ABC, the angle ABC = 90, Ba = BC. Point D is the midpoint of AB, CD, passing through point B, BC is the vertical CD, intersecting CD and Ca at points E and f respectively. The straight line passing through point a and perpendicular to AB intersects at point G, connecting DF. The following four conclusions are given: (1) Ag ratio AB = FG ratio FB; (2) point F is the midpoint of Ge; (3) AF = 3 molecular root 2Ab; (4) s triangle ABC = 5S triangle BDF, The correct conclusion is as follows There are specific solutions
• 10. As shown in the figure, in RT △ ABC, ∠ ABC is a right angle, ab = 3, BC = 4, P is the moving point on the edge of BC, let BP = x, if we can find a point Q on the edge of AC, so that ∠ BQP = 90 °, then the value range of X is______ ．
• 11. 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
• 12. 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
• 13. 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
• 14. As shown in the figure, in the triangle ABC, < a = 55 ° and H is the intersection of high BD and EC, then < BHC=
• 15. 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
• 16. 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
• 17. 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
• 18. If ∠ BDC = x + 2 / 3 ∠ a, find the degree of X?
• 19. 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
• 20. If ∠ BDC = α + 2 / 3 ∠ a, α is obtained