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

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

It is proved that: BA ⊥ AC, Da ⊥ AE, BAC = DAE = 90 ° in RT △ ABC and RT △ ade, BC = deab = ad, RT △ ade ≌ RT ≌ ABC, e = C, AC = AE, in △ ACM and △ AEN, C = EAC = AE, cam = ean ≌ ACM ≌ AEN (ASA), am = an

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

It is proved that: BA ⊥ AC, Da ⊥ AE, BAC = DAE = 90 ° in RT △ ABC and RT △ ade, BC = deab = ad, RT △ ade ≌ RT ≌ ABC, e = C, AC = AE, in △ ACM and △ AEN, C = EAC = AE, cam = ean ≌ ACM ≌ AEN (ASA), am = an

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

It is proved that: BA ⊥ AC, Da ⊥ AE, BAC = DAE = 90 ° in RT △ ABC and RT △ ade, BC = deab = ad, RT △ ade ≌ RT ≌ ABC, e = C, AC = AE, in △ ACM and △ AEN, C = EAC = AE, cam = ean ≌ ACM ≌ AEN (ASA), am = an

As shown in the figure, in ⊙ o, CD is the diameter, chord AB intersects CD at point m, and C is the midpoint of arc ACB, me ⊥ AC at point E, AC = 5, and CE ∶ EA = 3 ∶ 2
(1) The length (2) ⊙ o diameter of string ab
Second question

∵ Ce: EA = 3:2, AC = 5, ∵ CE = 2, AE = 3, ∵ C is the middle point of arc ACB, ∵ AC = BC, ∵ CD diameter, ∵ CD ⊥ AB, CE ⊥ C, ∵ RT Δ CMA ∵ RT Δ CEM, ∵ cm / CE = Ca / cm, cm ^ 2 = 10, CM = √ 10, ∵ am = √ (AC ^ 2-cm ^ 2) = √ 15, ∵ AB = 2am = 2 √ 15

As shown in the figure, in ⊙ o, CD is the diameter, chord AB intersects CD at point m, and C is the midpoint of arc ACB, me ⊥ AC at point E, AC = 5, and CE ∶ EA = 3 ∶ 2
Find the length of (1) string AB and the diameter of (2) ⊙ o

The drawing is not very standard. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; / & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; there is no need to do similar things here
I don't know if you learn similar triangles & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; / ∵ AC = 5 & nbsp; & nbsp; CE: AE = 3:2 & nbsp; E on AC
∵ AC = 5 & nbsp; & nbsp; CE: AE = 3:2 & nbsp; E on AC & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; CE = 3 & nbsp; AE = 2
The following is the 3 & nbsp; and it is the 3 & amp; nbsp; and it is the product, which is the product of the & nbsp; and the & amp & nbsp; and the & amp & nbsp; & amp & nbsp; & amp & nbsp; & amp & nbsp; and the & amp & nbsp; & amp & nbsp; and & amp & nbsp; & amp & amp & amp & amp & nbsp; & & amp & amp & nbsp; & amp & amp & nbsp; & amp & amp & nbsp; & amp & amp & amp & amp & amp & nbsp; & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & nbsp; & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & AMP & amp & amp & amp & amp & nbsp; & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & nbsp; & amp & amp & amp & amp & amp & amp & nbsp sp; / ∵ CD is the diameter & nbsp; & nbsp; C is the midpoint of the arc ACB
∵ CD is the diameter & nbsp; & nbsp; C is the midpoint of the arc ACB & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; I ∵ CD ⊥ AB & nbsp; CD bisects AB & nbsp;
⊥ CD ⊥ AB & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; if am = x & nbsp;, then cm = under (25-x & # 178;)
∵ - EAB + ∠ = 90 ° = - EAB + ∠ MCA & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Sub & nbsp; & nbsp; EM & # 178; = CM & # 178; - ce & # 178; = am & # 178; - AE & # 178;
This is the result of the following, which is the one of the case, which is the one of the following, which is the one of the Chinese, which is the part of the MCA, and the & amp & amp & nbsp; and the & nbsp; & amp & nbsp; & amp & nbsp; & amp & nbsp; & amp & nbsp; & amp & nbsp; & amp & nbsp; & nbsp; & & nbsp; & amp & nbsp; & nbsp; & & nbsp; & amp & amp & nbsp; & nbsp; & amp & nbsp; & amp & amp & amp & amp & amp & nbsp; & amp & amp & amp & amp & nbsp; & amp & amp & amp & nbsp; & amp & amp & amp & amp & amp & amp & amp & amp & amp & amp & nbsp; & & nbsp; & nbsp; & amp & amp & amp & amp & amp & AMP & amp & amp & nbsp; & amp & amp & amp & nbsp; nbsp; & nbsp; & & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & & nbsp; & & nbsp; & nbsp; & nbsp; & nbsp; sp; & nbsp; & nbsp; 25-x & # 178; - 9 = x & # 178; - 4
It's the point where you want to make your life, and it is the point where you want to make your life, and it is the point where you are the point where you are your & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & amp & nbsp; & & nbsp & nbsp; & nbsp & nbsp & nbsp; & nbsp & nbsp; & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; & nbsp & nbsp; & & nbsp & nbsp & nbsp & nbsp; & & nbsp; & & nbsp & nbsp & nbsp; & nbsp & & nbsp & nbsp & nbsp & & & nbsp & & nbsp & nbsp & nbsp & & nbsp & nbsp & nbsp & & nbsp & nbsp & & & nbsp & nbsp & nbsp & & nbsp & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp nbsp; & nbsp; & nbsp; & nbsp; x = √ 10
∴∠AEM=∠MEC                                        /               AM=√10 
∴△AEM∽△MEC                                       /         AB=2AM=2√10
AE:EM=EM:CE                                             /
EM=√6  AM=√10                                          /
AB=2AM=2√10                                            /

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

It is proved that: BA ⊥ AC, Da ⊥ AE, BAC = DAE = 90 ° in RT △ ABC and RT △ ade, BC = deab = ad, RT △ ade ≌ RT ≌ ABC, e = C, AC = AE, in △ ACM and △ AEN, C = EAC = AE, cam = ean ≌ ACM ≌ AEN (ASA), am = an

It is known that, as shown in the figure, ∠ BAE = ∠ CAD = 90 °, ad = AC, ab = AE, M is the midpoint of BC, and the extension line of AM intersects de at n
Verification: an ⊥ de
The picture may not come out
Old Tu can't come here. There's no way

Rotate as shown in the figure
Get & nbsp; parallelogram (BM = MC & nbsp; & nbsp; mam collinear & nbsp; proof omitted)
∠ cam + ∠ dam = 90 ° (top and bottom)
So am ⊥ Ma
Because & nbsp; Ma ‖ de & nbsp; (parallelogram)
So & nbsp; an ⊥ DC

As shown in the figure, △ ABC, Ag ⊥ BC is at point G, AB and AC are taken as one side to make rectangle ABME and rectangle acnf outside △ ABC respectively, and the 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 reason

He = HF. Reason: EP ⊥ GA and FQ ⊥ GA are made at crossing point E, and the perpendicular feet are p and q. ∵ quadrilateral. ABME is rectangular, ∵ BAE = 90 degree, ∵ bag + ∵ EAP = 90 degree, and ∵ Ag ⊥ BC, ∵ bag + ∵ ABG = 90 degree, ∵ ABG = ∵ EAP. ∵ AGB = ∵ EPA = 90 degree, ∵ ABG ∽ EAP, ∵ Ag: EP = AB: EA =AC: FA = k, Ag: EP = Ag: FQ. EP = FQ. In RT △ Eph and RT △ FQH, ∠ Eph = ∠ FQA ∠ EHP = ∠ fhqep = FQ, RT △ Eph ≌ RT △ FQH (AAS).. he = HF

It is known that in △ ABC, ab = AC = 5, BC = 8, point G is the center of gravity, then GA=______ ．

∵ AB = AC = 5, BC = 8, point G is the center of gravity, ∵ ad ⊥ BC, CD = 12bc = 12 × 8 = 4, ∵ ad = ac2 − CD2 = 25 − 16 = 3, ∵ GA = 2

In the isosceles triangle ABC, ab = AC, D is the middle point of the bottom edge BC, and the vertical bisector EF of AC edge intersects ad at point G
It is proved that GA = GB = GC, it is better to have graph

The distances from point F to the sides AB and AC of triangle ABC are equal,
The distance from point F to vertex a, B and C of triangle ABC is equal