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As shown in the figure, it is known that m and N are two points on the edge BC of △ ABC, and BM = Mn = NC is satisfied. A straight line parallel to AC intersects the extension line of AB, am and an at points D, e and f respectively. It is proved that EF = 3DE
It is proved that the parallel lines passing through N and m respectively intersect AB at h and G, NH crosses am at K,
∵BM=MN=NC，
∴BG=GH=HA，
Then HK = 1
2GM，GM=1
2HN，
∴HK=1
4hn, i.e. HK
KN=1
3，
DF ∥ HN,
∴DE
EF=HK
KN=1
3，
That is, EF = 3DE
As shown in the figure, m and N are two points on the edge BC of △ ABC, and BM = Mn = NC = am = an=______ .
∵BM=MN=NC=AM=AN，
﹤ amn is an equilateral triangle, ﹤ B = ∠ BAM,
∴∠MAN=∠AMN=60°．
∵∠B+∠BAM=∠AMN，
∴∠B+∠BAM=60°，
∴∠BAM=30°，
∴∠BAN=30°+60°=90°．
So the answer is: 90 degrees
In the acute triangle ABC, m and N are the points on AB and AC respectively, P is the point on Mn, BM / MA = an / NC = MP / PN, find: s triangle PBC / s triangle amn Sorry, missing a letter
Let BM / MA = an / NC = MP / PN = k, let BM / MA = K (BM + MA) / MA = 1 + K, MA / AB = 1 / (1 + k) BM / (BM + MA) = K / (1 + k) BM / AB = k / (1 + k) BM / AB = K / (1 + k) from an / NC = k an / (an + NC) = K / (1 + k) an / AC = K / (1 + k) (an + NC) / NC = 1 + K, NC / AC = 1 + K, NC / AC = 1 / (1 + k), from MP / PN = k, MP / (MP + PN) = K / (1 + k) MP / Mn = K / (1 + k) (1 + k) (k) (1 + k) (k) (k) (1 + k) (k) (k) (1 + k) (1 + k) (k) (1 + k) (k) (1 + k) (MP + PN) / PN = 1 + K pN / Mn = 1 / (1 + k) s triangle amn / s triangle ABC = (am * an) / (AB * AC) = (AM / AB) * (an / AC) = (1 / (1 + k)) * (K / (1 + k)) = K / (1 + k) ^ 2 ABN = (K / (1 + k) ^ 2) * s triangle ABC ABC s triangle BMP = (MP / Mn) * s triangle BMN = (MP / Mn) (BM / Mn) (BM / AB) * s triangle ABN = (MP / Mn) (BM / AB) (an / AC) * s triangle ABN = (K / (1 + k)) (K / (1 + k)) (K / (1 + k)) (K / (1 + k)) (K / (1 + k)) (K / (1 + k)) (k / (1 + k)) * s triangle ABC = (k ^ 3 / (1 + k) ^ 3) * s triangle ABC s triangle CNP = (PN / Mn) (NC / AC) * s triangle MNC = (PN / Mn) (NC / AC) * s triangle AMC (NC / AC) * s triangle AMC) AMC (NC / AC) * s triangle AMC) amc= (PN / Mn) (NC / AC) (AM / AB) * s triangle ABC =(1 / (1 + 1 + k)) (1 / (1 + K + k)) (1 / (1 + K + k)) (1 / (1 + K + k)) (1 / (1 + 1 + k) ^ 3) * s triangle ABC ABC = (1 / (1 + K + 2) / (1-k + K ^ 2) / (1-k + K ^ 2) / (1-k + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K ^ 2) / (1 + K + K + K + K ^ 2) / (1 + K + K + K + K ^ 2) + k) ^ 2) * s triangle ABC = (2k / (1 + K ^ 2)) * s triangle ABC So: s triangle PBC / s triangle amn = (2k / (1 + K ^ 2)) / (K / (1 + k) ^ 2) = 2
As shown in the figure, in RT △ ABC, ∠ BAC = 90 °, m and N are points on BC edge, BM = Mn = NC, if am = 4, an = 3, then Mn=______ .
As shown in the figure, AM2 = ad2 + MD2, an2 = AE2 + NE2, ad2 = 43, NE2 = 113, ∵ en is the median line of ⊥ CDM, so MD = 2NE, ∧ no ⊥ Mo, MD ⊥ ed, quadrilateral Oden is parallel four
In the triangle ABC, ab = AC point m, n is on BC and am = an, please explain the reason why BM = NC
Proof: do AF ⊥ BC
Because AB = AC, AF ⊥ BC
Three lines in one, f is the midpoint of BC
BF=CF
Similarly, because am = an, AF ⊥ Mn
Three lines in one, f is the midpoint of Mn
MF=NF
BF-MF=CF-NF
BM=CN
As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connect an, cm, intersect at point P, try to find the degree of ∠ APM
As shown in the figure, make the vertical line of AB through a, and intercept AK = CN = MB, connect km and KC, then
Because am = BC, AK = BM, ∠ Kam = ∠ B = 90 °,
So △ Kam ≌ △ MBC,
So km = cm, ∠ AMK = ∠ MCB
Because ∠ CMB + ∠ MCB = 90 °,
So ∠ CMB + ∠ AMK = 90 °
So ∠ KMC = 90 °
Therefore, △ KMC is an isosceles right triangle, ∠ MCK = 45 °
Because ∠ Kam = ∠ B = 90 °, AK = CN,
So AK ∥ CN,
So anck is a parallelogram,
So KC ‖ an,
So ∠ APM = ∠ KCM = 45 °
As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connect an, cm, intersect at point P, try to find the degree of ∠ APM
As shown in the figure, make the vertical line of AB through a, and intercept AK = CN = MB, connect km and KC, then
Because am = BC, AK = BM, ∠ Kam = ∠ B = 90 °,
So △ Kam ≌ △ MBC,
So km = cm, ∠ AMK = ∠ MCB
Because ∠ CMB + ∠ MCB = 90 °,
So ∠ CMB + ∠ AMK = 90 °
So ∠ KMC = 90 °
Therefore, △ KMC is an isosceles right triangle, ∠ MCK = 45 °
Because ∠ Kam = ∠ B = 90 °, AK = CN,
So AK ∥ CN,
So anck is a parallelogram,
So KC ‖ an,
So ∠ APM = ∠ KCM = 45 °
In the right angle △ ABC, ∠ B = 90 degrees, point m on AB, so that am = BC, point n on BC, so that CN = BM, connect cm, an, intersect at point P, find ∠ APM
It is easy to know △ CNQ ≌ △ BMC ≌ BMC ≌ ≌ ≌ ≌ ≌ ≌ ≌ ≌ ≌ ≌ degrees, AQ = NQ, then
As shown in the figure, in △ ABC, ∠ B = 90 °, M is the point on AB, so that am = BC, n is the point on BC, so that CN = BM, connect an, cm, intersect at point P, and prove ∠ APM = 4 APM = 45
As shown in the figure, the vertical line of AB is made by passing through a, and AK = CN = MB, connecting km, KC, then because am = BC, AK = BM, ∠ Kam = ∠ B = 90 °, so △ Kam ≌ △ MBC, KM = cm, ∠ AMK = ∠ MCB because ∠ CMB + ∠ MCB = 90 ° so ∠ KMC = 90 ° so △ KMC is isosceles right triangle, ∠
In △ ABC, ab = AC, points m and N are on BC, and am = an, please specify BM = CN Fig.: the top angle is A, the left side is 「 B, and the right side is 「 C
Because am = an
Therefore, amn = anm
So ∠ AMB = ∠ ANC
Because AB = AC
So ∠ B = ∠ C
So △ ABM ≌ △ ACN (AAS)
So BM = CN
RELATED INFORMATIONS
- 1. As shown in the figure, △ ABC, ab = AC, points m and N are on the straight line where BC is located, and am = an Excuse me: BM = CN? Please state the reasons
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