As shown in the figure, in △ ABC, ∠ C = 90 °, ad bisects ∠ BAC, ab = 5, CD = 2, then the area of △ abd is () A. 2B. 5C. 10D. 20

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• 1. As shown in the figure, in △ ABC, ∠ C = 90 °, CD ⊥ AB, CM bisects AB, CE bisects ∠ DCM, then the degree of ∠ ace is______ ．
• 2. As shown in the figure, in △ ABC, ab = AC, point E is on the extension line of Ca, and ED ⊥ BC is on D
• 3. In the triangle ABC, if the bisector of the inner angle B and C intersects at point E and the bisector of the outer angle intersects at point F, and angle a is equal to 70 degrees, then angle BEC equals? Angle BFC equals?
• 4. It is known that in the triangle ABC, ∠ B = 1 / 2 ∠ a, CD is perpendicular to BC, CE is the middle line on the edge BD, and AC = 2 / 1ab is proved Using the properties of right triangle (1)
• 5. It is known that be is the bisector of the inner angle of triangle ABC and CE is the bisector of the outer angle of triangle ABC Add a reason after each step
• 6. As shown in the figure, in the triangle ABC, the angle a = 60 degrees. The bisectors BD and CE of the triangle ABC intersect at point F. if be = 4 and CD = 2, find the length of BC
• 7. In triangle ABC, bisectors be and CD of angle ABC and angle ACB intersect at point O. if angle a = 60 °, what is the relationship between BC, CE and BD
• 8. Two bisectors BD and CE of triangle ABC intersect at point O, angle a = 60 degrees, proving: CD + be = BC There is an answer to this question on the Internet, but I can't understand it because of my low IQ. Dashen explains why ∠ EOB = 1 / 2 (180 ° - 60 °) = 60 ° This is a certain method: take a point F on the edge of BC, make BF = be, link of ∵ BD is the angular bisector, BF = be, Bo is the common edge, ∴△BEO≌△BFO →∠EOB＝∠FOB＝∠COD ∵∠A＝60° The angle bisector is BD and CE ∵∠EOB＝1/2（180°－60°）＝60° Then ∠ COF = 180 ° - ∠ FOB - ∠ cod = 60 ° And ∵ CE is the angular bisector and CO is the common edge, ∠ COF = ∠ cod = 60 ° (proved) ∴△COF≌△COD →CF＝CD ∴BC＝BF＋CF＝BE＋CD Or there are other ways,
• 9. In triangle ABC, BD and CE are the bisectors of angle B and angle C respectively. Given be = CD, it is proved that triangle ABC is isosceles triangle I'm sorry, but I can imagine everything
• 10. In RT △ ABC, ∠ B = 90 ° is connected to the midpoint m of AC side, and BM = cm = am is proved
• 11. In the triangle ABC, point O is a moving point on the side of AC, and a straight line Mn / / BC is made through point O. let the bisector of the intersection angle BCA of Mn be at point E, and the bisector of the outer angle of the intersection angle BCA be at point E Line and point F. (1) explore the quantitative relationship between line OE and of and prove it. (2) when point O moves on edge AC, will quadrilateral BCFE be a diamond? If so, please prove it. If not, explain the reason. (3) when point O moves to where, and what conditions does triangle ABC meet, quadrangular aecf is a square?
• 12. As shown in the figure, △ ABC, point O is a moving point on the edge of AC. make a straight line Mn ‖ BC through point O. let the bisector of the intersection angle BCA of Mn be at point E, and the bisector of the external angle ACD of △ ABC be at point F (1) Try to explain the line segment EO = fo (2) Guess: when point O moves to where, the quadrilateral aecf is a rectangle? And explain the reason
• 13. In △ ABC, point O is a moving point on edge AC, and a straight line Mn parallel to BC is made through O. let Mn intersect the bisector of ∠ BCA at point e. the bisector of the outer angle of ∠ BCA intersects with point F 1. When point O moves on edge AC, will the quadrilateral BCFE be a diamond? If so, please prove. If not, explain the reason; 2. When point O moves to where, and △ ABC satisfies what conditions, the quadrilateral aecf is a square?
• 14. In △ ABC, point P is a moving point on edge AC, and a straight line Mn ∥ BC is made through point P. let Mn intersect ∥ BCA and its outer angle bisector be at point E, and intersect ∥ BCA and its outer angle be at point F (1) Verification: PE = PF (2) Can the quadrilateral BCFE be a diamond when point P moves on edge AC? Explain why (3) If there is a point P on the side of AC, let the quadrilateral aecf be a square, and AP / BC = (√ 3) / 2
• 15. As shown in the figure, in the triangle ABC, point O is a moving point on the edge of AC, and a straight line Mn is parallel to BC through point O. suppose that the bisector of the intersection angle BCA of Mn is at point E, and the bisector of the external angle BCA of Mn is at point F. if CE = 12, CF = 5, the length of OC can be obtained
• 16. As shown in the figure, in the triangle ABC, point O is a moving point on the side of AC. a straight line Mn is parallel to BC through point O, and the bisector of Mn ∠ BCA is set at point E, The bisector of the outer angle of intersection ∠ BCA is at point F (1) try to explain: EO = fo; (2) when point 0 moves to where, the quadrilateral aecf is a rectangle? And explain the reason
• 17. As shown in the figure, in △ ABC & nbsp;, point O is a moving point on the edge of AC, and a straight line Mn ‖ BC is made through point O. suppose that the angular bisector of Mn intersecting ∠ BCA is at point E, and the outer angular bisector of intersecting ∠ BCA is at point F. (1) try to explain EO = fo; (2) when point O moves to where, is the quadrilateral aecf rectangular? (3) when the point O moves to where, and what condition does △ ABC satisfy, the quadrilateral aecf is a square? And explain the reason
• 18. In the triangle ABC, point O is a moving point on the side of AC. when passing through point O, a straight line Mn is parallel to BC. Let the bisector of the intersection angle BCA of Mn be at e, and the bisector of the intersection angle BCA be at E Bisector at point F. (1) proof; EO = fo, (2) when point O moves to where, quadrilateral aecf is rectangular, prove your conclusion
• 19. It is known that in △ ABC, D is the midpoint of AB, e is the point on AC, EF ‖ AB, DF ‖ be. (1) conjecture: the relationship between DF and AE is______ Try to explain the correctness of your conjecture
• 20. In the RT triangle ABC, the angle B = 90 ° ED is the vertical bisector of AC, intersecting BC and AC at points E and D respectively, connecting AE, if the angle BAE: the angle BAC = 1:5