As shown in the figure, it is known that CD is perpendicular to ab at point D, be is perpendicular to AC at point E, and CD intersects be at point o (1) If OC = ob, prove that point O is on the bisector of angular BAC (2) If point O is on the bisector of angle BAC, OC = ob is proved

Analysis: ① connect Ao. Prove △ CEO ≌ △ BDO through the decision theorem ASA of congruent triangles, and then know OC = ob according to the equality of corresponding sides of congruent triangles;
② From the properties of angular bisector, OD = OE can be obtained, and then △ DOB ≌ △ EOC can be proved, OB = OC can be proved
① Connect Ao
∵CD⊥AB,BE⊥AC,
∴∠CEB=∠BDO=90°；
And ∵ ∠ COE = ∠ BOD (equal to vertex angle),
The remaining angles of equal angles are equal;
In △ CEO and △ BDO,
∠ C = ∠ B, OC = ob (known), ∠ COE = ∠ EOD & nbsp; & nbsp; & nbsp;
∴△CEO≌△BDO（ASA）,
Ψ OE = OD (the corresponding sides of congruent triangles are equal),
The point O is on the bisector of BAC;
② It is proved that: Ao bisection ∠ BAC, CD ⊥ AB, be ⊥ AC,
∴OD=OE,
In △ DOB and △ EOC,
∠DOB=∠EOC,OD=OE,∠ODB=∠OEC,
∴△DOB≌△EOC（ASA）,
∴OB=OC．

In known triangle ABC, ab = AC, ad ⊥ BC in D, de ⊥ AC in E, f is the midpoint of de

It is proved that: let be intersect with AD at m, intersect with AF at n ∵ AB = AC, ad ⊥ BC ∫ BD = CD = 1 / 2BC ≁ de ⊥ AC ∪ C + ∠ CDE = ∠ CDE + ade = 90 ∪ C = ∠ ade, and then: ∪ CED = ∠ ADC = 90 °, C = ∠ C ∪ CDE ∪ CAD ∪ AD / CD = de / CE ∫ de = 2DF, CD = 1 / 2BC ∪ AD / (1 / 2BC) = (2DF) / CE ∪ (2ad) /

In the triangle ABC, ab = AC, ad is the height on the edge of BC, De is perpendicular to AC and E, f is the midpoint of ED
In the triangle ABC, ab = AC, ad is the height on the edge of BC, De is perpendicular to AC and E, f is the midpoint of ED
It is proved that AF is perpendicular to be
How to prove this problem? This problem has no picture, so I need to draw my own picture,

Easy to know DC / ad = EC / de
Because D is the midpoint of BC and F is the midpoint of de
So BC = 2dc, de = 2DF
So BC / ad = EC / DF
And because angle ADF = angle BCE
So triangle BCE is similar to triangle ADF
So DAF = CBE
So AF is perpendicular to be

In △ ABC, ab = AC, ∠ a = 120 °, the vertical bisector of AB intersects at point m, AB intersects at point E, the vertical bisector of AC intersects at point n, AC intersects at point F
Verification: BM = Mn = NC

Because AB = AC, ∠ a = 120 °, so, ∠ B = ∠ C = 30 °. EM and FN are the vertical bisectors of AB and AC respectively, so am = BM, an = BN, ∠ BAM = ∠ B = 30 °, can = ∠ C = 30 °, so, ∠ man = 120 ° - 30 ° - 30 ° = 60 °, and ∠ amn = ∠ B + ∠ BAM = 30 ° + 30 ° = 60 °, so

As shown in the figure, in isosceles △ ABC, P is a moving point on the bottom edge, m and N are the midpoint of AB and BC respectively. If the minimum value of PM + PN is 2, then the perimeter of △ ABC is?
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Let o be the midpoint of AC
The minimum value of PM + PN is 2
∴OM+ON=2
∵ △ ABC is an isosceles triangle, ∠ ABC = 120 degree
∴AB=BC,
OM=ON=AM=BM=BN=CN=1
∴AB=BC=2
AC²=AB²+BC²-2AB*BC*cos120°
AC=2√3
The circumference of ABC: 2 + 2 + 2 √ 3 = 4 + 2 √ 3

In the triangle ABC, ab = AC, BAC = 120 degrees, m and N are the middle points of AB and AC respectively, M is the first moving point of BC at the bottom, and the minimum value of PM + PN is 2
Find the perimeter of the triangle ABC.

1. It should be a moving point of bottom BC?
2. The key to solve this problem is to prove that P is the midpoint of BC
3. If the connection Mn is made, the connection Mn is parallel to the bottom BC
4. Take the bottom BC as the symmetry axis to make the symmetry point m 'of point m, and the line M'n intersects with BC at point P, then PN + PM is the minimum. (this is a theorem of primary school. There are two villages AB on one side of the river L. we need to build a pool C by the river to supply water to the two villages. Ask where C is built, AC + BC is the minimum. This is the method used.)
5. Mm 'intersects with BC at point O, and PQ intersects Mn at point q.op = 1 / 2Mn, PQ = om = 1 / 2m M', indicating that point P is the midpoint of M'n and Q is the midpoint of Mn. Then AQ is perpendicular to Mn, and AP is perpendicular to BC. Because triangle ABC is isosceles triangle, point P is the midpoint of edge BC
6. MP = NP = 1 / 2Ac = 1 / 2Ab = 1, then AC = AB = 2, BC = 2 √ 3

It is known that in isosceles ABC, AC = 2, P is a moving point on the bottom AC, and m n is the midpoint of AB and BC respectively. If the minimum value of PM + PN is 2, what triangle is △ ABC? Please explain the reason and find its perimeter

Equilateral triangle with perimeter of 6
Reason: if PM + PN is the shortest, then p is on the middle perpendicular of Mn and intersects AC with P, so AP = CP = 1. Similarly, it can be proved that AP = NP = 1 and Mn is the median of triangle, so Mn = 1, so MPN is equilateral triangle, so

It is known that in △ ABC, ∠ ACB = 90 ° AC = BC, the straight line Mn passes through point C, and ad ⊥ Mn is in D, be ⊥ Mn is in E

It is proved that: ① ∵ - ACB = 90 °, be ⊥ CE, ad ⊥ CE, ∵ - BEC = - ACB = - ADC = 90 °, ∵ - ACE + - BCE = 90 °, ∵ BCE + - CBE = 90 °, ∵ ACD = - CBE, in △ ADC and △ CEB, ≌ - ADC = - BEC ≌ ACD = - cbeac = BC, ≌ - ADC ≌ - CEB (AAS)

As shown in the figure, in △ ABC, be and CD are bisectors of ∠ ABC and ∠ ACB respectively, am and an are perpendicular to CD and be respectively, and the perpendicular foot is m and N. verification: Mn / / BC

Extending the extension line of am an BC at point p q
According to CD be is angular bisector and CD vertical AP be vertical aq
It can be concluded that △ AMC ≌ PMC △ anb ≌ QNB
Then m n is the midpoint of AP aq
Mn is the median of △ Apq
Mn / / PQ is Mn / / BC

As shown in the figure, be and CF are angular bisectors of △ ABC, an ⊥ be is in N, am ⊥ CF is in M, proving: Mn ∥ BC

It is proved that the intersection of an and am at points D and g. ∵ be is the angular bisector of ∠ ABC, be ⊥ AG, ∵ BAM = ∠ BGM, ∥ ABG is the isosceles triangle, and ∥ BM is also the middle line of isosceles triangle, that is am = GM. similarly, an = dn, ∥ Mn is the median line of △ ADG, ∥ Mn ∥ BC

As shown in the figure, it is known that CD is perpendicular to ab at point D, be is perpendicular to AC at point E, and CD intersects be at point o (1) If OC = ob, prove that point O is on the bisector of angular BAC (2) If point O is on the bisector of angle BAC, OC = ob is proved