As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on the segments AB, BC and Ca respectively, (1) If ad = be = CF, is △ def an equilateral triangle? Try to prove your conclusion; (2) If △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion

As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on the segments AB, BC and Ca respectively, (1) If ad = be = CF, is △ def an equilateral triangle? Try to prove your conclusion; (2) If △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion

(1) The proof is as follows: ∵ ABC is an equilateral triangle,  a = ∠ B = ∠ C, ab = BC = Ca, and ∵ ad = be = CF, ∵ DB = EC = FA, (2 points) ∵ ADF ≌ △ bed ≌ △ CFE, (3 points)

Δ ABC is the inscribed triangle of circle O, ab = AC, e is a point on AC arc in circle O, the extension line of BC and AE intersects point D, connects CE, and ab × CE = AE × CD

Because AB = AC, so ∠ ABC = ∠ ACB
Because ∠ ABC + ∠ AEC = 180 °∠ ACB + ∠ ACD = 180 °, so ∠ AEC = ∠ ACD
Because ∠ ACB = ∠ CAD + ∠ ADC ∠ ABC = ∠ CED = ∠ CAD + ∠ ace
So ∠ ace = ∠ ADC
So △ AEC is similar to △ ACD
AE / AC = CE / CD, so AC × CE = AE × CD, that is ab × CE = AE × CD

As shown in the figure, in the triangle ABC, angle c = 90 degrees, ad bisector angle cab intersects CB at D, CD = 3, BD = 5, find the length of AD

Ideas:
The vertical line AC is made by D, and the vertical foot is e
The equal distance between the bisector and the two sides proves that ACD is equal to AED
Then, be edge is obtained by Pythagorean theorem
Let AC = AE = X
In triangle ABC
From the Pythagorean equation we can get

In the right angle trapezoid ABCD, ad ∥ BC, ∠ ABC = 90 °, BD ⊥ DC, be = DC, CE bisection ∠ BCD, crossing AB at point E, the following conclusions are obtained ① BH = DH; ② ch = (radical 2 + 1) eh ③ s △ enh / s △ EBH = eh / EC It is a high school entrance examination question in Wuhan, Hubei Province,

The correct ones are (2,3)
1. When h is passed, HM is perpendicular to BC to m, and because CE bisects ∠ BCD, BD is perpendicular to CD ℅ HD = HM, and HM is not equal to Hb
2. Let BM be y and en be X
∵ ABC = 90 °, BD ⊥ DC, BD = DC, CE bisection ∠ BCD
∴BM=HM=y=DH,BH=√2y
And ∵ ABC = 90 
∴∠ABD=45°
∴EN=BN=x
And ∵ easy to prove ∵ Anh ∵ CDH
∴NH=√2y-x
CD=BD=√2y+y
∴（√2y-x)/y = x/(√2y+y)
The solution is y = x, and then ch: eh = Y / (√ 2y-y) = √ 2 + 1
3. The area of the triangle enh can be obtained by x = y in 2: the area of triangle EBH = [x (√ 2x-x)] / 2: [x (√ 2x-x) + x 2] / 2 = (2 - √ 2) / 2
Because ch = (radical 2 + 1) eh  eh: EC = (2 - √ 2) / 2
☆⌒_ I hope I can help you~

As shown in the figure, in trapezoid ABCD, ad ∥ BC, ∠ ABC = 60 ° BD bisection ∠ ABC, BC = 2Ab

It is proved that the vertical lines of BC are drawn through two points a and D respectively, and they intersect points E and F,
∴∠AEF=∠DFE=90°，
∵AD∥CB，
∴∠DAE=∠AEF=∠DFE=90°，
The quadrilateral aefd is a rectangle,
∴AD=EF，
∵ BD bisection ∵ ABC,
∴∠ABD=∠DBC=30°，
∵AD∥CB，
∴∠ADB=∠DBC=∠ABD，
∴AB=AD，
∴EF=AD=AB，
∵BC=2AB，
∴BE+FC=AB．
Be = FC = 1 can be seen from ∠ Abe = 60 °
2AB
By proving △ Abe ≌ △ DCF, ab = DC

It is known that in the triangle ABC, BD is the bisector of ∠ ABC, de ‖ BC intersects AB with E, EF ‖ AC intersects BC with F, then be = EC, why? BE=FC

prove:
∵ BD bisection ∵ ABC
∴∠ABD＝∠CBD
∵DE∥BC
∴∠EDB＝∠CBD
∴∠ABD＝∠EDB
∴BE＝DE
∵DE∥BC,EF∥AC
The parallelogram cdef
∴DE＝FC
∴BE＝FC

In the triangle ABC, ad bisects ∠ BAC, CE ⊥ ad in O and AB in E, EF ∥ BC. It is proved that EC bisection ∠ def

∵ EF||BC
∴∠FEC=∠DCE
∵∠BAD=∠DAC,CE⊥AD
∴ EG=CG
∴ ∠DEC=∠DCE
∴ ∠DEC=∠FEC

As shown in the figure, in △ ABC, the point E is on AC, the point n is on BC, and a point F is found on AB to minimize the circumference of △ enf, and the reasons are given

Make the symmetry point e 'of e with respect to AB, connect e' n, intersect with ab at point F, connect EF, NF, at this time, the circumference of △ enf is the smallest,
The reason is: the perimeter of △ EFN = EF + FN + en, and en is a fixed value,
By using the shortest segment between two points, we get that EF + FN = e ′ F + FN = e ′ n is the smallest, then the perimeter is the smallest

Taizhou, 2012) as shown in the figure, it is known that the straight line L is separated from ⊙ O, OA ⊥ L is at point A, OA=5.OA intersects ⊙ O at point P, and AB is tangent to ⊙ O at point B

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CD is the height on the hypotenuse ab of RT △ ABC, if the ratio of lengths of two right angles AC, BC is AC: BC = 3:4 Find (1) ad: BD; (2) if AB = 25cm, find the length of CD

(1) : 3:4 (2) (50 / 7) times root 3
The first question is solved by triangle similarity, which is actually a theorem. The second question is that triangle CDB is similar to triangle ADC