In the graph + equilateral triangle ABC, if D, e and F are on BC, AC and ab respectively, and CD = BF, BD = CE, then what degree is the angle EDF

In the graph + equilateral triangle ABC, if D, e and F are on BC, AC and ab respectively, and CD = BF, BD = CE, then what degree is the angle EDF

Because CD = BF, BD = CE, angle B = angle c = 60 degrees, so triangle BDF is equal to triangle CED, so angle BFD = angle CDE, and because angle B + angle BFD = angle FDE + angle CDE = angle CDF, angle EDF = 60 degrees

As shown in the figure, AB is the diameter of semicircle o, ab = 2. Rays am and BN are tangent lines of semicircle. Take a point D on am, connect BD, intersect semicircle with point C, and connect AC Ad = 1, and DPQ is the tangent of the semicircle ᚉ OP = 1 and op ⊥ DP ᚉ DQ ∥ AB (please ask... Why op ⊥ DP, DQ is ∥ AB?

 BF / ob = AB / ad

As shown in the figure, the diameter of ⊙ o is ab = 2, am and BN are its two tangent lines, de cuts ⊙ o in E, intersects am in D, intersects BN in C. let ad = x, BC = y (1) Confirmation: am ‖ BN; (2) Find the relation between Y and X; (3) Find the area s of quadrilateral ABCD and prove that s ≥ 2

(1) It is proved that: ∵ AB is the diameter, am and BN are tangent,

As shown in the figure, points B and D are on ray am, points c and E are on ray an, and ab = BC = CD = de. if ∠ a = 18 °, then the degree of ∠ EDM is

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In diamond ABCD, m and N are points on the edge of BC and CD respectively. If am = an = Mn = AB, try to find the degree of angle C

Because am = an = Mn, so the angle amn = angle MNA = angle man = 60 degrees, and because the quadrilateral ABCD is diamond, so angle B = angle D, angle bad = angle c, ab = ad, so AB = ad = am = an, so angle ABM = angle BNA = angle and = angle NDA, let angle ABM be x, then x of 360-4x = 180-x-60 solution is 80, so angle bad = angle c = 180-80 = 100
That is, the angle c is 100 degrees

In diamond ABCD, m and N are points on the edge of BC and CD respectively. If am = an = Mn = AB, find the degree of ∠ C

Since ABCD is diamond shaped, AB=BC=CD=AD
Known: am = an = Mn = ab
Then, △ amn is equilateral triangle, △ ABM and △ adn are isosceles triangle
Let ∠ B = ∠ d = X
Then, ∠ AMB = ∠ and = X
Therefore, ∠ BAM = ∠ Dan = 180 ° - 2x
Then, ∠ bad = 2 * (180 ° - 2x) + 60 ° = 420 ° - 4x
Because AB / / CD
Therefore, ∠ bad + ∠ ADC = 180 °
That is, (420 ° - 4x) + x = 180 °
===> 420°-3x=180°
===> x=80°
Therefore, ∠ C = 180 ° - ∠ d = 180 ° - 80 ° = 100 °

As shown in the figure, ab = ad, BC = De, and BA ⊥ AC, Da ⊥ AE, Can you prove that am = an?

Proof: BA ⊥ AC, Da ⊥ AE,
∴∠BAC=∠DAE=90°，
In RT △ ABC and RT △ ade,
BC＝DE
AB＝AD ，
∴Rt△ADE≌Rt△ABC，
∴∠E=∠C，AC=AE，
In △ ACM and △ AEN,
∠C＝∠E
AC＝AE
∠CAM＝∠EAN
∴△ACM≌△AEN（ASA），
∴AM=AN．

As shown in the figure, ⊙ o, CD is the diameter, chord AB intersects BC at point m, C is the midpoint of arc ACB, me is perpendicular to e, AC = 5, and Ce: EA = 3:2, am / AC = AE / am 1. Find the length of string AB; 2. Find the diameter of ⊙ o;

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As shown in the figure, ab = ad, BC = De, and BA ⊥ AC, Da ⊥ AE, Can you prove that am = an?

Proof: BA ⊥ AC, Da ⊥ AE,
∴∠BAC=∠DAE=90°，
In RT △ ABC and RT △ ade,
BC＝DE
AB＝AD ，
∴Rt△ADE≌Rt△ABC，
∴∠E=∠C，AC=AE，
In △ ACM and △ AEN,
∠C＝∠E
AC＝AE
∠CAM＝∠EAN
∴△ACM≌△AEN（ASA），
∴AM=AN．

As shown in the figure, in △ ABC, Ag ⊥ BC is at point G, with AB and AC as the sides, rectangle ABME and rectangular acnf are made to △ ABC respectively. Ray GA intersects EF at point h. If AB = KAE, AC = Kaf, try to explore the quantitative relationship between he and HF, and explain the reasons

He = HF. Reason: the crossing point E is EP ⊥ GA, FQ ⊥ GA, and the vertical feet are p and q. ∵ the quadrilateral ABME is a rectangle, ? BAE = 90 °, bag + ∠ EAP = 90 ° and ∵ Ag ⊥ BC, ∵ bag + ∠ ABG = 90 °