As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons

As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons

The △ AFC is an isosceles triangle
In △ bad and △ BCE,
 B = ∠ B (common angle), ∠ bad = ∠ BCE, BD = be,
∴△BAD≌△BCE（AAS），
∴BA=BC，∠BAD=∠BCE，
∴∠BAC=∠BCA，
Ψ BAC - ∠ bad = ∠ BCA - ∠ BCE, i.e., ∠ fac = ∠ FCA
∴AF=CF，
The △ AFC is an isosceles triangle

As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons

Δ AFC is an isosceles triangle. The reasons are as follows: in △ bad and △ BCE, ? B = ∠ B (common angle), ∠ bad = ∠ BCE, BD = be,  bad ≌ △ BCE (AAS), ? Ba = BC, ? bad = ≌ BCE, ? BAC = ≌≌△ BCE, i.e., ﹤ fac = ∠ FCA. ﹥ AF = CF

As shown in the figure, in the triangle ABC, point E is on AB, point D is on BC, BD = be, angle bad = angle BCE, ad and CE are compared with F, to judge the shape of triangle AFC To be clear, this problem is not very good, please look at it! It's question 5 on page 15 of the math exercise book (I'm not sure how to draw a picture. Excuse me.) Please hurry up Have a look at the exercise books Good + 10 points, help me solve, I thank you Be quick

It's an isosceles triangle
In △ BCE and △ bad: BD = be, ∠ BCE = ∠ bad, ∠ B is the common angle, ≌△ bad
ν Ba = BC, that is, △ ABC is an isosceles triangle
∴∠BCA=∠BAC
∴∠FCA=∠FAC
So FC = FA

As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons

The △ AFC is an isosceles triangle
In △ bad and △ BCE,
 B = ∠ B (common angle), ∠ bad = ∠ BCE, BD = be,
∴△BAD≌△BCE（AAS），
∴BA=BC，∠BAD=∠BCE，
∴∠BAC=∠BCA，
Ψ BAC - ∠ bad = ∠ BCA - ∠ BCE, i.e., ∠ fac = ∠ FCA
∴AF=CF，
The △ AFC is an isosceles triangle

As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = EF, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons

∵BD=BE,
∠BAD=∠BCE,
Is the common angle
∴△ABD≌△CBE.
∴AB=BC.
From this we can get
△AEF≌△CDF.
∴AF=CF.
The △ AFC is an isosceles triangle

As shown in the figure, in the triangle ABC, angle a = angle c, point D and then AB, point E is on the extension line of CB, and angle e = angle BDE

It is proved that: the extended ed ed and AC intersection point P is the △ EPC ∠ C + ∠ e + e + ∠ EPC = 180 ° in Δepc, the △ DAP ∠ a + ∠ ADP + ∠ APD = 180 ° in Δdap, the Δdap ∠ a + ∠ ADP + APD = 180 °, C + ∠ e + e + ∠ ? e = e = 1 ∵ EDB, ∵ EDB, ∵ EDB, ? EDB, ? a = a = C ? C ? C = 180 °νdp

As shown in the figure, in the triangle ABC, ab = AC, ad is perpendicular to CB to D, e is any point on ad, EM is perpendicular to AB to m, en is perpendicular to AC to N. please prove that EM = en What conclusion can be conjectured by proving EM = en

prove:
∵AB=AC,AD⊥BC
∴∠MAE=∠NAE
∵∠AME=∠ANE=90°,AE=AE
∴△AME≌△ANE
∴EM=EN
The conclusion is that the distance from any point on the high line on the base of isosceles triangle to two waist is equal

As shown in the figure, in △ ABC and △ DCB, ab = DC, AC = dB, AC and DB intersect at point M (1) Verification: △ ABC ≌ △ DCB; (2) Confirmation: BM = cm

It is proved that: (1) ∵ AB = DC, AC = dB, BC = CB,
∴△ABC≌△DCB（SSS）．
(2) Method 1:
∵△ABC≌△DCB，
∴∠1=∠2，
∴BM=CM．
The second method of proof is as follows
∵△ABC≌△DCB，
∴∠A=∠D，
And ∵ AB = DC, ∵ 3 = ∠ 4,
∴△ABM≌△DCM（AAS），
∴BM=CM．

As shown in the figure, we know that in the right triangle ABC, ∠ ABC = 90 ° and ab = ad, CB = CE, try to find the degree of ∠ EBD. (please write the solution process clearly)

Let ∠ a = x °,
∵∠ABC=90°，
∴∠C=（90-x）°，
∵AB=AD，CE=CB，
∴∠ABD=∠ADB，∠BEC=∠EBC，
∴∠ADB=（180−x
2）°=（90-x
2）°，∠EBC=[180-（90-x）]÷2=[45+x
2]°，
∴∠DBC=∠ADB-∠C=（90-x
2）°-（90-x）°=（x
2）°，
∴∠EBD=∠EBC-∠DBC=（45+x
2）°-（x
2）°=45°．

As shown in the figure, center O is the circumscribed circle of △ ABC, CG is the diameter, CE is perpendicular to e, CA = 4, CB = 6, CE = 3. Find the length of CG

In the right triangle CBE, CE = 3, hypotenuse BC = 6, so the angle ABC = 30 degrees, the angle AGC = the angle ABC = 30 degrees (equal to the circular angle of the arc). Because CG is the diameter of the circle, the vertical AG of Ca is obtained, so CG = 2Ac = 2 * 4 = 8