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In the triangle ABC, D is the midpoint of BC and DM is perpendicular to DN. If BM + CN = DM + DN, verify Ad = 1 / 4 (AB + AC)
prove:
Extend nd to e so that de = DN, connect be
∵BD=CD,∠BDE=∠CDN
∴⊿BDE≌⊿CDN（SAS）
∴BE=CN,∠EBD=∠C
∵DM⊥DN
∴∠MDE=90º
Connecting me
Then me? 2 = DM? 2 + de? 2 = DM? 2 + DN? 2
∵BM²+CN²=BM²+BE²=DM²+DN²
∴BM²+BE²=ME²
∴∠MBE=90º
∴∠ABC+∠EBD=∠ABC+∠C=90º
∴∠BAC=90º
ad = half BC [the center line of the hypotenuse of a right triangle is equal to half of the hypotenuse]
∴AD²=¼BC²
∵BC²=AB²+AC²
∴AD²=¼（AB²+AC²）
As shown in the figure, △ ABC is an equilateral triangle, and BM = CN, am and BN intersect at point P, then ∠ APN=______ .
In △ ABM and △ BCN,
AB＝BC
∠ABM＝∠BCN
BM＝CN ，
∴△ABM≌△BCN，∴∠BAM=∠CBN
∵∠APN=∠ABN+∠BAM，∠ABN+∠CBN=60°
∴∠APN=∠ABC=60°，
So the answer is 60 degrees
It is known that △ ABC is an equilateral triangle, points m and N are on ray BC and ray CA respectively, and BM = CN. If BN and am intersect at point P, the degree of angle BPM is calculated
BC=AB,
∠NCB=∠MBA=60°
CN=MB,
△BCN≌△ABM,[SAS]
∠N=∠M;
∠ pan = ∠ Mac, [opposite vertex angle]
∠BPM=∠N+∠PAN=∠M+∠MAC=∠ACB=60°.
(1) As shown in the figure, in positive △ ABC, point m and point n are points on BC and Ca respectively, and BM = CN. Connect am and BN. The two lines intersect at point Q, and calculate the degree of ∠ aqn (2) Change the "positive △ ABC" in question 1 into square ABCD, regular pentagon ABCDE, regular hexagon ABCDEF , n-polygon ABCD N. The other conditions remain unchanged. According to the solution idea of question 1, the degree of ∠ aqn is deduced respectively, and the conclusion is filled in the following table: ⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙ Regular polygon square regular pentagon regular hexagon Regular n-polygon The degree of ∠ aqn
(1) In △ ABM and △ BCN, ab = BC ∠ ABC = ∠ C = 60 ° BM = CN, ABM ≌ △ BCN (SAS), BAM = ∠ NBC, aqn = ∠ BAM + ∠ ABQ, NBC + ∠ ABQ, ABQ, ABM = 60 ° aqn = 60 °; (2) from (1), we can see that ∠ aqn = the size of an angle of each polygon
It is known that the parallelogram ABCD, ad = a, ab = B, ∠ ABC = α. The point F is the point on the line BC (except the end point B and C), connecting AF, AC, connecting DF, and extending the extension line of DF intersection AB at point E, connecting CE (1) When f is the midpoint of BC, it is proved that the areas of △ EFC and △ ABF are equal; (2) When f is any point on BC, are the areas of △ EFC and △ ABF equal? Give reasons
(1) It is proved that: ∵ point F is the middle point of BC, ᙽ BF = CF = 12bc = A2, and
Given that AC ‖ de and AC = De, CE intersect point B, AF, DG are the midlines of the triangle ABC triangle BDE, it is proved that the quadrilateral agdf is a parallelogram
Because AC / / De, and AC = De, so ACDE is a parallelogram. So CE and AD are diagonals of parallelogram, intersection point is their bisector, triangle ABC and triangle BDE are congruent triangles
It is known that AC is a diagonal of the parallelogram ABCD, BM ⊥ AC, DN ⊥ AC, and the perpendicular feet are m and N respectively, Verification: quadrilateral bmdn is parallelogram
Proof: BM ⊥ AC, DN ⊥ AC,
∴∠DNA=∠BMC=90°，
∴DN∥BM，
∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC，AD=BC，
∴∠DAN=∠BCM，
∴△ADN≌△CBM，
∴DN=BM，
The quadrilateral bmdn is a parallelogram
As shown in the figure, AF and be are bisected, EC and DF are bisected. It is proved that the quadrilateral ABCD is a parallelogram The teacher said it Prove: connect Mn (AF has 2 intersections with EC, connect) Because BM = em, en = CN So 2Mn is equal to and parallel to BC Similarly, 2Mn is equal to and parallel to AD So one in two 1BC is equal to and parallel to one-half ad So BC is equal to and parallel to AD So the quadrilateral ABCD is a parallelogram
Connect AE, De, BF, CF, EF
∵BP=EP,AP=FP
The quadrilateral affe is a parallelogram,
∵DQ=FQ,EQ=CQ
The quadrilateral DCFE is a parallelogram,
∴AB=EF=CD
AB‖EF‖DC
The quadrilateral ABCD is a parallelogram
I also just thought of it! And thank you very much for the hint from the next one!
In the parallelogram ABCD, AE bisection angle bad intersects BC at point E, BF bisection angle ABC intersects ad at point F. it is proved that AF is perpendicular to BF
The problem should be to prove AE ⊥ BF ∵ AE BF bisection ∵ bad ∵ ABC and quadrilateral ABCD is a parallelogram ? ad ≁ BC ? bad + ABC = 180 ° let AE and BF intersect m ᚉ BAM + ∠ AMB = 1 / 2 ∠ ABC + 1 / 2 ? bad = 1 / 2 * 180 ° = 90 ° in △ ABM, it is obtained that: AMB = 90
As shown in the figure, in the parallelogram ABCD, the bisector of ∠ bad intersects with edge BC at point E, and bisector of ∠ ABC intersects with edge ad at point F. please prove that the abef of quadrilateral is diamond
It is proved that: ∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC，
∴∠4=∠5，
The bisector BF of ∵ ABC,
∴∠3=∠4，
∴∠3=∠5，
∴AF=AB，
∵AD∥BC，
∴∠1=∠AEB，
The bisector AE of ∵ BAC,
∴∠1=∠2，
∴∠2=∠AEB，
∴BE=AB，
∴AF=BE，
∵AF∥BE，
The quadrilateral abef is a parallelogram,
∵AF=AB，
The parallelogram abef is a diamond
RELATED INFORMATIONS
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