It is known that, as shown in the figure, angle ABC = angle DCB, BD and Ca are the bisectors of angle ABC and angle DCB respectively
Connect ad ∵ BD and Ca to divide ∵ ABC and ∵ DCB ∵ ABO = ∵ OBC, ∵ DCO = ∵ OCB ∵ ABC = ∵ DCB ≌ ABC - ≌ ABO = ≌ DCB - ≌ DCO, that is
RELATED INFORMATIONS
- 1. E. B and C are on the same straight line, Ba bisects ∠ EBD, ∠ DBC = 30 ° and calculates the degree of ∠ ABC
- 2. In △ ABC, Ba bisects ∠ DBC, ∠ BAC = 118 ° BD ⊥ AC in D, and calculates the degree of ∠ C
- 3. As shown in the figure, point D is a point on the AB extension line of the equilateral triangle ABC. Take CD as the side length, make the equilateral triangle CDE, and calculate the degree of the angle EBD
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- 12. As shown in the figure, the known points E and C are on the line BF, be = CF, ab ‖ De, ∠ ACB = ∠ F
- 13. As shown in the figure, given that AD and be are the heights of △ ABC, ad and be intersect at point F, and ad = BD, can you find the congruent triangle in the figure? If you can find it, please explain why
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- 17. As shown in the figure: O is any point in △ ABC, a '. B'. C 'is OA.OB.OC Is triangle ABC similar to triangle a'b'c '? Why?
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