Take the right angle side of the RT triangle ABC as the diameter, make a semicircle o, intersect the oblique side with D, OE parallel with AC, and intersect AB with e. it is proved that De is the tangent of circle o

Connect o, D
∵OE‖AC=〉∠ODC=∠DOE
∵ OC, od are the radius of circle o = ｛ ODC = ∠ OCD
∵180°-∠ODC=∠ODC+∠OCD=2∠DOC
∵180°-∠ODC=∠DOE+∠EOB
=〉∠DOE+∠EOB=2∠DOC=〉∠EOB=∠DOC=〉∠EOB=∠DOC=∠DOE
And ∵ EOB = ∵ doe
∵ ob, OD is the radius of circle o = 〉 ob = OD
OE=OE=〉△OBE≌△ODE=〉∠0DE=∠OBE
And ∵ ∵ ABC is a right triangle = 〉 ab ⊥ BC = 〉 OBE = 90 degrees
=〉∠0DE=∠OBE=90°
=ED is perpendicular to OD
=De is the tangent of circle o
I don't know if you have the symbols. My teacher does. If the school is different, the teacher will teach differently. = > it's a symbol to push out. I'm afraid to learn differently

In the triangle ABC, the angle a = 60 degrees, BC is the fixed length, and the circle O with BC as its diameter intersects AB and AC at points D and e respectively to connect de and OE The following conclusions are as follows: 1. BC = 2DE; 2. The distance between D and OE remains unchanged; 3. BD + CE = 2DE; 4. OE is the tangent line of the circumscribed circle of the triangle ade. Which is correct? This question is a little difficult, the answers are 1, 2, 4. Please write down the reasons in detail

1. Let ∠ ABC = x, ∠ ACB = y, x + y = 120, and because od = ob, OE = OC, so ∠ ODB = ∠ ABC, ∠ OEC = ∠ ACB

As shown in the figure, in △ ABC, the bisectors AD and CE of ∠ B = 60 °, BAC, ∠ ACB intersect at point O, which shows the reason why AE + CD = AC

It is proved that AF = AE is selected on AC and connected with of,
Then △ AEO ≌ △ AFO (SAS),
∴∠AOE=∠AOF；
∵ AD and CE are equally divided into ∵ BAC, ∵ ACB,
∴∠ECA+∠DAC=1
2（180°-∠B）=60°，
Then ∠ AOC = 180 ° - ∠ ECA - ∠ DAC = 120 °;
Ψ AOC = ∠ DOE = 120 °, AOE = ∠ cod = ∠ AOF = 60 °, (equal to vertex angle)
Then ∠ COF = 60 °,
∴∠COD=∠COF，
And ? FCO = DCO, CO = Co,
∴△FOC≌△DOC（ASA），
∴DC=FC，
∵AC=AF+FC，
∴AC=AE+CD．

In the triangle ABC, ab = AC, ad vertical BD, AE vertical CE, and ad = AE, BD and CE intersect with point O. please explain the reason why ob = OC Eight class navigation, mathematics Zhejiang Education Press, B version 16 pages

As shown in the figure, we can prove congruence with HL of right triangle, if ∠ ADB = ∠ AEC = 90 °, hypotenuse AB = AC, right angle side ad = AE, so △ ADB is all equal to △ AEC, so ∠ abd = ∠ ace, because AB = AC, so ∠ ABC = ∠ ACB, so ∠ ABC - ∠ abd = ∠ ACB, that is ∠ OBC = ∠ OCB, so ob = OC

In RT △ ABC, ∠ a = 90 ° CE is an angular bisector, which intersects with high ad at F, and FG ‖ BC intersects AB with G, Confirmation: AE = BG

Make eh ⊥ BC at h, as shown in the figure,
∵ e is the point on the bisector, eh ⊥ BC, EA ⊥ ca,
∴EA=EH，
∵ ad is the height of ᙽ ABC, EC is equal to ∵ ACD,
∴∠ADC=90°，∠ACE=∠ECB，
∴∠B=∠DAC，
∵∠AEC=∠B+∠ECB，
∴∠AEC=∠DAC+∠ECA=∠AFE，
∴AE=AF，
∴EH=AF，
∵FG∥BC，
∴∠AGF=∠B，
In △ AFG and △ EHB,
∠GAF＝∠BEH
∠AGF＝∠B
AF＝EH ，
∴△AFG≌△EHB（AAS）
∴AG=EB，
AE + eg = BG + Ge,
∴AE=BG．

In RT △ ABC, ∠ a = 90 ° CE is an angular bisector, which intersects with high ad at F, and FG ‖ BC intersects AB with G, Confirmation: AE = BG

Make eh ⊥ BC in H, as shown in the figure, ∵ e is the point on the bisector, eh ⊥ BC, EA ⊥ Ca, ᙽ EA = eh, ∵ ad is the height of ⊥ ABC, EC bisection ∠ ACD, ∠ ADC = 90 °, ACE = ∠ ECB, ∵ B = ∠ DAC, ? AEC = ∠ DAC + ∠ ECA = ∠ AE = AF, ﹤ eh = AF

As shown in the figure, in right angle trapezoid ABCD, ad ∥ BC, ∠ ABC = 90 °, de ⊥ AC at point F, intersection BC at point G, extension line crossing AB at point E, and AE = AC (1) Results: BG = FG; (2) If ad = DC = 2, find the length of ab

(1) It is proved that: connect AG, ∵ ABC = 90 °, de ⊥ AC at point F, ∵ ABC = ∠ AFE. In △ ABC and △ AFE,  ABC ≌△ AFE (AAS), ab = AF. in RT △ ABG and RT △ AFG, Ag = AgaB = AF ≌ RT ≌ RT △ AFG (HL)

In RT △ ABC, ∠ a = 90 ° CE is an angular bisector, which intersects with high ad at F, and FG ‖ BC intersects AB with G, Confirmation: AE = BG

In the triangle ABC, AB is equal to 15, AC is equal to 13, and height ad is equal to 12. What is the circumference of the triangle ABC

You can draw the triangle on the paper first because ad ⊥ BC
From the Pythagorean theorem, CD ^ 2 = AC ^ 2-ad ^ 2, that is, CD = 5
Similarly, BD ^ 2 = AB ^ 2-ad ^ 2, that is, BD = 9
So BC = BD + CD = 5 + 9 = 14
So the circumference of the triangle is 14 + 15 + 13 = 42

In the triangle ABC, given that AB is equal to 15, AC is equal to 13, and height ad is equal to 12, what is the circumference of the triangle ABC

∵ad⊥bc
From Pythagorean theorem
: CD 2 = AC 2 - ad ^ 2, i.e. CD = 5
In the same way:
BD? 2 = AB? Ad? 2, that is, BD = 9
∴bc=bd+cd=5+9=14
The circumference of the triangle = 14 + 15 + 13 = 42

Take the right angle side of the RT triangle ABC as the diameter, make a semicircle o, intersect the oblique side with D, OE parallel with AC, and intersect AB with e. it is proved that De is the tangent of circle o