As shown in the figure, △ ABC, D is on AB, ad = BD = CD, de ∥ AC, DF ∥ BC

As shown in the figure, △ ABC, D is on AB, ad = BD = CD, de ∥ AC, DF ∥ BC

De / / AC DF / / BC just prove that ∠ C is a right angle
AD=BD=CD
CD is the middle line of △ ABC, △ ADC △ DBC is isosceles △ B = ∠ BCD ∠ a = ∠ ACD
∠B+∠BCD+∠A+∠ACD=180
2∠B+2∠A=90
∠C=90

As shown in the figure, point D is a point on the hypotenuse ab of RT △ ABC, de ⊥ BC is in E, DF ⊥ AC is in F, if AF = 15, be = 10, then the area of quadrilateral decf is______ ．

If ∵ DF ⊥ AC, de ⊥ BC, ∵ DFC = ≁ C = ≁ Dec = 90 ° and ∵ quadrilateral dfce is a rectangle, it is easy to know that DF ∥ BC, then ∵ ADF = ≁ B, AFD = ≁ DEB, ∩ ADF ∽ DBE, ∩ dfbe = afde, i.e. de · DF = AF · be = 150, ∩ s rectangular dfce = de · DF = 150, i.e. the area of quadrilateral dfce is 150

As shown in the figure, in △ ABC, ∠ ACB = 90 ° and ∠ BAC = 30 °, both △ Abe and △ ACD are equilateral triangles, BF = Fe, DF and AC intersect at point m, proving that am = MC

It is proved that: as shown in the figure, AF, FC, ∵ Abe are equilateral triangles, BF = EF, ∵ AF is bisector of ∠ BAE, ∵ BAF = ∠ BAE = 12 × 60 ° = 30 °, ∵ BAC = 30 ° and ≌ BAF = ∠ BAC = 30 ° in △ Abf and △ ABC, ≌ BAF = ∠ BAC ∠ AFB = ∠ ACB = 90 ° AB = AB, ≌

In a right triangle, ∠ ABC = 90 ° with ab as the diameter, make ⊙ o intersection AB at point D, e is the midpoint of BC connecting de. prove that the line De is the tangent of ⊙ o

It is proved that ∠ ABC = 90 ° and ⊙ o with ab as the diameter should intersect AC at point D. because OA = OD, so ∠ oad = ∠ ODA, because OE / / AC, so ∠ EOB = ∠ oad, so ∠ EOBO = EO, OB = od = R, so △ EBO congruent △ EOD, so ∠ Edo =}