As shown in the figure, in △ ABC, AE is the angular bisector, and ∠ B = 52 °, C = 78 ° to find the degree of ∠ AEB

As shown in the figure, in △ ABC, AE is the angular bisector, and ∠ B = 52 °, C = 78 ° to find the degree of ∠ AEB

In △ ABC, ∵ B = 52 °, ∵ C = 78 °, ∵ BAC = 180 ° - 52 ° - 78 ° = 50 °, ∵ AE is the angular bisector, ∵ BAE = 12 ∠ BAC = 12 × 50 ° = 25 °, ∵ AEB = 180 ° - 52 ° - 25 ° = 103 °

As shown in the figure, in △ ABC, AE is the angular bisector, and ∠ B = 52 °, C = 78 ° to find the degree of ∠ AEB

In △ ABC, ∵ B = 52 °, ∵ C = 78 °, ∵ BAC = 180 ° - 52 ° - 78 ° = 50 °, ∵ AE is the angular bisector, ∵ BAE = 12 ∠ BAC = 12 × 50 ° = 25 °, ∵ AEB = 180 ° - 52 ° - 25 ° = 103 °

As shown in the figure, in △ ABC, EF is the median line of △ ABC, D is a point on the side of BC (not coincident with B and C), ad and EF intersect at point O to connect de and DF. To make quadrilateral AEDF a parallelogram, conditions need to be added______ (only add one condition, the answer is not unique)

Add: BD = CD. Reason: ∵ EF is the median line of △ ABC ∵ CF = AF, AE = 12ab. ∵ BD = CD, ∵ point D is the midpoint of BC, DF is the median line. ∥ DF ∥ AE therefore, in order to make quadrilateral AEDF parallelogram, according to a group of parallel and equal quadrilateral is parallelogram, we need to add the condition BD = CD. So the answer is BD = CD (the answer is not unique)

It is known that: as shown in the figure, in △ ABC, De is the median line, EF ‖ AB, EF intersects BC at point F. verification: F is the midpoint of BC

It is proved that: as shown in the figure, ∵ in △ ABC, De is the median line, point E is the midpoint of AC, and ∵ EF ∥ AB, ∥ EF is the median line of △ ABC, point F is the midpoint of BC

In triangle ABC, ab = AC, angle BCA = 48 degrees, CECF is divided into three equal parts ∠ ACB, intersecting ad to EF, connecting be and extending AC to point G, connecting FG, then ∠ AGF=

Is ad high?

As shown in the figure, ad, be and CF are the three bisectors of △ ABC, then: ∠ 1 + 2 + 3=______ ．

∵ ad, be and CF are the three bisectors of △ ABC, ∵ 1 = 12 ∵ BAC, ∵ 2 = 12 ∵ ABC, ∵ 3 = 12 ∵ ACB, ∵ 1 + 2 + 3 = 12 (∵ BAC + ABC + ACB),