In the triangle ABC, a = 65, D, e and F are the points on BC, Ca and ab respectively, and BF = DF =, CE = de. the degree of EDF is calculated

In the triangle ABC, a = 65, D, e and F are the points on BC, Ca and ab respectively, and BF = DF =, CE = de. the degree of EDF is calculated

Solution: ∵ a = 65 °, then ∵ B + ∵ C = 115 ° ∵ BF = DF ∵ B = ∵ BDF (the two straight lines are equal, and the corresponding angles are also equal) ce = de ∵ d = ∵ EDC (ditto) ∵ BDF + ∵ EDC = ∵ B + ∵ C = 115 ∵ EDF = 180 ° - 115 ° = 65 °

It is known that in △ ABC, ab = AC, point D is on AC, and BD = BC= AD.DE Is the midline of △ abd, f is the point on AB, and DF = BF, find the degree of ∠ EDF

It is known that: C = ABC = 2 A, so a = 36 ° C = ABC = 72 ° because BD = BC = ad, so De is isosceles △ abd midline, which is also high, that is, in the isosceles triangle BDC with ade = 90 ° it is easy to get that DBC = 180 ° - 2x72 ° = 36 ° so FBD = 72 ° - 36 ° = 36 ° from DF = BF

Known: as shown in the figure, in △ ABC, ab = AC, point D is on AC, and BD = BC = ad, De is the middle line of △ abd, f is a point on AB, and DF = BF, find ∠ EDF

∵ in the triangle ABC, ab = AC, point D is on AC, and BD = BC = ad,
The angle a = 36 ° (the angle a = x ° can be set to get x + 2x + 2x = 180)
And ∵ BF = DF, ∵ DFE = 36 ° + 36 ° = 72 °
And De is the middle line of triangle abd, and BD = ad
So De is perpendicular to ab
The angle EDF is 90-72 = 18 degrees

In the triangle ABC, ad is the angle bisector, De is perpendicular to ab at point E, DF is perpendicular to AC at point F, and the relationship between AD and EF is judged

Triangle ade and ADF congruence
So AE = AF
Let AD and EF intersect at o
Then the triangle AOE and AOF are congruent
So ad and EF are vertical