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As shown in the figure, in the RT triangle ABC, the angle c = 90 degrees, the DB bisector angle ABC intersects AC at point D, De is the vertical bisector of AB and intersects AB at point E (1) Find the degree of angle A (2) If BC = 6, AC = 8, then the circumference of the triangle BDC is
In
In the RT triangle ABC, the angle ACB = 90, the vertical bisector EF of side AC intersects AC at point E, AB at point F, BG perpendicular to AB, crossing EF with point G It is proved that CF is the middle term of the ratio between EF and FG
It is proved that: because EF ⊥ AC, BC ⊥ AC, so EF / / BC, and AE = AC, so AF = FB, that is, f is the midpoint of AB, so CF = AB / 2 = AF = FB in △ AEF and △ GBF, ∠ AEF = ∠ GBF = 90 degrees and ∠ AFE = ∠ GFB (equal angle), so △ AEF ≈ GBF, so AF / EF = GF / BF, namely af * FB = EF * GF
As shown in the figure, in RT △ ABC, the vertical bisector De of ∠ ACB = 90 ° intersects with the extension line of BC at F. if ∠ f = 30 ° de = 1, the length of EF is () A. 3 B. 2 C. Three D. 1
Connect AF,
∵ the vertical bisector De of AB intersects with the extension line of BC at F,
∴AF=BF，
∵FD⊥AB，
∴∠AFD=∠BFD=30°，∠B=∠FAB=90°-30°=60°，
∵∠ACB=90°，
∴∠BAC=30°，∠FAC=60°-30°=30°，
∵DE=1，
∴AE=2DE=2，
∵∠FAE=∠AFD=30°，
∴EF=AE=2，
Therefore, B
As shown in the figure, in RT △ ABC, the vertical bisector ed of ∠ C = 90 ° AB intersects BC at point D, and ∠ CAD: ∠ cab = 1:3, calculate the size of ∠ B
∵ De is the vertical bisector of ab;
∴AD=DB;
∴∠DAB=∠B;
CAD: ∠ cab = 1:3
Ψ DAB: ∠ cab = 2:3, that is ∠ B = 2 / 3 ∠ cab;
And ∠ B + ∠ cab = 90 °;
∴∠B=36°
In △ ABC, ∠ C = 90 °, the vertical bisector De of segment AB intersects BC with D, and the perpendicular foot is e. if ∠ cab = 65 °, then ∠ CAD ＝
40°
As shown in the figure, in △ ABC, the bisector ad of ∠ cab intersects with the vertical bisector De of BC at D, DM ⊥ AB at m, and DN ⊥ AC extension line at n. try to explain: BM = CN Can't use Pythagorean theorem, root sign and trigonometric function!
Connecting BD and CD
∵ ad bisection ∵ cab DM ⊥ AB in M, DN ⊥ AC in n
ν DM = DN, and ∠ DMB = ∠ DNC = 90 °
∵ de vertical bisector BC ᙽ DB = DC
∴Rt⊿DMB≌Rt⊿DNC﹙HL﹚
∴BM=CN
It is known that in △ ABC, the bisector ad of ∠ cab and the vertical bisector De of BC intersect at point D, DM ⊥ AB and m, DN ⊥ AC intersect the extension of AC at n. what do you think is the relationship between BM and CN? Try to prove your findings
BM=CN．
Reason: connect BD, CD,
∵ ad bisection ∵ BAC, DM ⊥ AB, DN ⊥ AC,
∴DM=DN，
∵ de bisects BC vertically,
∴BD=CD，
In RT △ BMD and RT △ CND
A kind of
BD＝CD
DM＝DN
∴Rt△BDM≌Rt△CDN（HL），
∴BM=CN．
In △ ABC, the bisector ad of cab and the vertical bisector De of BC intersect at point d DM ⊥ AB at point m, DN ⊥ AC at point n. It is proved that BM = CN
Connect BD, CD
Because ad bisects ∠ cab, DM ⊥ AB, DN ⊥ AC, DM = DN;
Because De is the vertical bisector of BC, BD = CD
Therefore, right triangle BDM, congruent right triangle CDN (HL),
Therefore, BM = CN
It is known that in △ ABC, the bisector ad of ∠ cab and the vertical bisector De of BC intersect at point D, DM ⊥ AB and m, DN ⊥ AC intersect the extension of AC at n. what do you think is the relationship between BM and CN? Try to prove your findings
BM=CN．
Reason: connect BD, CD,
∵ ad bisection ∵ BAC, DM ⊥ AB, DN ⊥ AC,
∴DM=DN，
∵ de bisects BC vertically,
∴BD=CD，
In RT △ BMD and RT △ CND
A kind of
BD＝CD
DM＝DN
∴Rt△BDM≌Rt△CDN（HL），
∴BM=CN．
It is known that: as shown in Fig. 12-39, in △ ABC, ab = AC, ∠ ABC = 30 °, the vertical bisector of segment AB intersects the extension line of Ca respectively, and CB is at points D and E It was proved that de = 2be
Proof: connect AE
∵AB=AC,∠ABC=30°
∴∠B=∠C=30°
∴∠DAB=60°
∵DE⊥AB
∴∠D=90°﹣∠DAB=30°
∵ De is the vertical bisector of ab
∴BE=AE
∴∠BAE=∠B=30°
∴∠DAE=∠DAB﹢∠BAE=90°
∴DE=2AE
∴DE=2BE
RELATED INFORMATIONS
- 1. It is known that the triangle ABC, CA = CB, point O is on the vertical bisector of Ca and CB, and m.n is on a straight line AC.BC Go ahead, It is proved that CN + Mn = am
- 2. In △ ABC, ∠ a = 1 2∠B=1 3 ∠ ACB, CD is the height of △ ABC, CE is the angular bisector of ∠ ACB, and the degree of ∠ DCE is calculated
- 3. As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC
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- 5. As shown in the figure, in diamond ABCD, AE ⊥ BC, perpendicular foot is e, angle DAE = 2, angle BAE, BD = 15, find the length of AC, ab
- 6. As shown in the figure, in △ ABC, D is a point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
- 7. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
- 8. As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad
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- 10. As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De
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- 14. In △ ABC, ab = BC, ∠ B = 90 °, M is the midpoint of AC, D.E is the moving point on AB and BC respectively, and BD = CE, the shape of △ DEM is obtained and proved
- 15. As shown in the figure, in the isosceles triangle ABC, ab = AC, D is a point on the extension line of AB, BD = AB, CE is the center line on waist AB, and verify; CD = 2ce needs specific process
- 16. As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion
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