In the triangle ABC, ab = AC, D is a point on BC, and

In the triangle ABC, ab = AC, D is a point on BC, and

Make a vertical line of BC through f e. the perpendicular foot is m n
Triangle DCF similar triangle DBE
therefore
ne：mf=db：dc
ne×dc=mf×db
1/2ne×dc=1/2mf×db
S triangle BDF = s triangle CDE

It is known that in the triangle ABC, ∠ a = 90 °, ab = AC, and D is the midpoint of BC. 1) e and F are the points on AB and AC respectively, and there are still be = AF

When point E does not coincide with point B, or point F does not coincide with point C
Because the triangle ABC is an isosceles right triangle
So DH = DG = AB / 2

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 90 °, BD is the bisector of ∠ ABC, CE ⊥ BD, the perpendicular is e, and the extension of Ba and CE intersects at point F. (1) find the triangle congruent with △ abd in the figure, and give the reason for congruence; (2) explain BD = 2ec; (3) if AB = 5, find the length of AD

It is proved that: (1) in △ abd ≌ △ ACF. ≌ ≌ AB = AC, ∠ BAC = 90 °, ≌ fac = ∠ BAC = 90 °, ≔ BD ⊥ CE, ∠ BAC = 90 °, ≌ ADB = ∠ EDC, ≌ abd = ∠ ACF, ≂ in △ abd and △ ACF, ≌ bad = ∠ cafab = AC ∠ ADB = ∠ ACF, ≌ abd ≌ ACF (ASA), ≌ (2) ≌ a

It is known that be and CF are in triangle ABC

First, make the auxiliary line, extend an and am to BC, and the intersection points are g and h respectively
∵ cn is the bisector of ∠ C
∴∠ACN=∠GCN
And ∵ an ⊥ cn
∴∠ANC=∠GNC
Cn is the public side
∴△ACN≌△GCN
∴AN=GN
N is the midpoint of Ag
Similarly, M is the midpoint of ah
In △ AGH, Mn ‖ GH
∴MN∥BC

As shown in the figure, it is known that be and CF are bisectors of angle B and angle C in triangular ABC, am is perpendicular to be, an is perpendicular to be, CF is perpendicular to N, Mn is parallel to BC

What do you want to ask? If it is to prove that Mn is parallel to BC Extending am and an intersect BC at Q, P ∠ FCB ＋ NPQ = 90 degree, ∠ EBC ＋ mqp = 90 degree. From the angle bisector, ∠ MBC ＋ Apq = 90 degree, we know that ∠ FCB ＋ AQP = 90 degree, there is upper, ∠ AQP = ∠ Apq. Triangle Apq is isosceles

As shown in the figure, in △ ABC, ab = AC, points F and E are the points above AB and AC respectively, am ⊥ CF is at point m, an ⊥ be is at point n, and am = an, the verification is: △ Abe ≌ △ ACF

It is proved that: am ⊥ CF, an ⊥ be ≌ BNAC = ∠ CMA = 90 ° in RT △ ABN and RT △ ACM, ab = acan = am, ≌ RT ≌ ABN ≌ RT △ ACM (HL), ≌ Abe = ∠ ACF, in △ Abe and △ ACF, ≌ Abe = acfab = AC ≌ BAE = CAF, ≌ Abe ≌ ACF (ASA)

As shown in the figure, in △ ABC, ab = AC, BD, CE are bisectors of the angle, and an ⊥ CE is at point M

It is estimated that the original question is:
In △ ABC, ab = AC, BD, CE are bisectors of the angle, an ⊥ CE at point n, am ⊥ BD at point M
Verification: am = an
It is proved that: ab = AC
∴∠ABC=∠ACB.
BD and CE are bisectors
In addition, ABM = ACN, AMB = ANC = 90 degrees, ab = AC
∴⊿ABM≌⊿ACN(AAS),AM=AN.

The copy task of a school was originally undertaken by a copy agency. The relationship between the charge y (yuan) and the number of copy pages x (pages) is as follows: X (pages) 100 200 400 1000 Y (yuan) 40 80 160 400
(1) If y and X meet a certain functional relationship learned in junior high school, find the analytic formula of the function; (2) now B copy agency says: if the school pays a contract fee of 200 yuan per month, it can charge 0.15 yuan per page. Then the relationship between Y (yuan) and X (pages) is______ (3) draw the function images in (1) and (2) in the given coordinate system, and answer which copy agency should be chosen when the number of pages copied per month is about 1200?

(1) Let the analytic formula be y = KX + B, substituting (100, 40), (200, 80) into 100k + B = 40200k + B = 80, the solution is k = 0.4b = 0, so y = 0.4x (x > 0 and is an integer); (2) the functional relationship between the monthly charge y (yuan) of B copy agency and the number of copy pages x (pages) is y = 0.15x + 200 (x ≥ 0 and is an integer); (3) the graph is as follows, from the graph, we can see that the number of copy pages per month is about 1200 Right should choose B copy agency