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Circle O is the circumscribed circle of triangle ABC, chord cm is perpendicular to AB, CN is the diameter, f is the midpoint of arc ab. the bisection angle NCM of CF is proved
F is the midpoint of arc ab
So ∠ ACF = ∠ BCF
And ∠ B = ∠ n
Cm crossed AB with E
∠BEC=∠NAC=90°
∴∠ACN=∠BCE
So CF bisection angle NCM
Circle O is the circumscribed circle of triangle ABC, chord cm is perpendicular to AB, CN is the diameter, f is the midpoint of arc ab. it is proved that arc an = arc BM
The intersection of an, cm and ab is marked as D
∵ cn is the diameter cm ⊥ ab
﹤ can = 90 degrees ﹤ CDB = 90 degrees
∵ ANC, ABC are the circular angles of - AC
∴ ∠ANC = ∠ABC
∵ ∠ACN = 180-∠CAN-∠ANC =180-∠CDB-∠ABC
∴∠ACN = ∠BCM
⌒AN=⌒BM
As shown in the figure, in △ ABC, ab = ad = DC, ∠ bad = 36 ° and calculate the degrees of ∠ B and ∠ C
AB=AD,∠BAD=36°
∴∠B=∠ADB=72°
Ad = DC
∴∠C=∠DAC
And ∠ C + ∠ DAC = ∠ ADB = 72 °
∴∠C=36°
In △ ABC, ad equals dB and EDB equals DAC. It is proved that △ ABC is similar to △ EAD
It is known that: as shown in the figure, △ ABC, ad = dB, ∠ EDB = ∠ DAC. Verification: △ ABC ∽ EAD. Examination point: Determination of similar triangle. Special topic: proof question. Analysis: according to the judgment of similar triangle, we should carefully examine the problem and choose the appropriate judgment method. Proof: ∵ ad = dB,
∴∠B=∠BAD．
∵∠BDA=∠EDB+∠C=∠DAC+∠ADE,
∴∠C=∠ADE．
∴△ABC∽△EAD．
As shown in the figure, in △ ABC, D and E are the two points on BC respectively, ∠ B = ∠ EAC, ∠ ADC = ∠ DAC Ad bisection ∠ BAE
∵∠ADC=∠B+∠BAD，
∴∠DAC=∠EAC+∠DAE．
And ? ADC = ∠ DAC, B = ∠ EAC,
∴∠BAD=∠DAE，
Ψ ad bisection ∠ BAE
Known: as shown in the figure, ∠ DAC = ∠ B, verification: ∠ ADC = ∠ BAC
It is proved that: ∵ DAC = ∠ B, ∠ C = ∠ C,
∴∠ADC=180°-∠C-∠DAC，∠B=180°-∠C-∠BAC，
∴∠ADC=∠BAC．
D is a point on edge BC of the triangle ABC,
Because angle ADC = angle bad + angle B, angle BAC = angle bad + angle DAC, because,
Known: as shown in the figure, ∠ DAC = ∠ B, verification: ∠ ADC = ∠ BAC
It is proved that: ∵ DAC = ∠ B, ∠ C = ∠ C,
∴∠ADC=180°-∠C-∠DAC，∠B=180°-∠C-∠BAC，
∴∠ADC=∠BAC．
In the triangle ABC, ab = BC, ad is the center line on the BC edge, e is a point on the extension line of BC, and CE = CB
Extend ad to f so that DF = ad
Then triangle abd is similar to FCD
∠ABC=∠BCF、CF=AB=CE
∠ACF=∠ACE( ∠ACE =∠BAC+∠B=∠ACB+∠FCB=∠ACF)
AC=AC
So AFC is equal to AEC
So angle DAC = angle CAE
As shown in the figure: any straight line passing through the vertex C of △ ABC is crossed with edge AB and midline ad at f and e respectively. It is proved that AE: ed = 2AF: FB
Because DM / / CF, the triangle AEF is similar to the triangle ADM, so AE: ad = AF: am, so AE * am = af * ad, so AE * (AF + FM) = af * (AE + ed), AE * FM = af * ed, because ad is the middle line, so D is the middle point, and because DM / / CF, M is the midpoint of BF, so FB = 2fm, AE * FB = AE * 2fm = 2AF * e
RELATED INFORMATIONS
- 1. The vertex C passing through △ ABC as a straight line intersects with edge AB and median line ad at points F and e respectively. It is proved that AE / de = 2AF / BF
- 2. O is a point in the equilateral triangle ABC, and the over current o is OD ∥ AB intersecting BC at point D and OE ∥ AC crossing BC at point E, what triangle ode is?
- 3. (1) As shown in Figure 1, ab = BC = Ca, OD and OE are radius of ⊙ o, OD ⊥ BC is at point F, OE ⊥ AC is at point G, It is proved that the area of the shadow quadrilateral ofcg is 1 of the area of △ ABC 3． (2) As shown in Fig. 2, if ∠ DOE keeps 120 ° angle unchanged, It is proved that when ∠ DOE rotates around point O, the area of the figure (shadow part in the figure) surrounded by two radii and two edges of △ ABC is always 1 of the area of △ ABC 3．
- 4. It is known that in △ ABC, D is the midpoint of AB, e is a point on AC, AE = 2ce, CD, be intersect at o point, OE = 2cm. Find the length of Bo
- 5. In the triangle ABC, ab = AC, and the midpoint of BC is d. draw an equilateral triangle def so that the vertices E and F are parallel to BC on sides AB and AC respectively
- 6. As shown in the figure, in △ ABC, ∠ BAC = 90 °, ab = AC, point D is on BC, and BD = Ba, point E is on the extension line of BC, and CE = Ca, try to find the degree of ∠ DAE (1) If the condition of "ab = AC" is removed and the other conditions remain unchanged, will the degree of ∠ DAE change?
- 7. As shown in the figure, in the triangle ABC, the angle c = 90 degrees, ad is the bisector of angle a, ab = 26, AC = 10cm, and find the distance from D to ab From top to bottom, from left to right is ABCD
- 8. In RT △ ABC, ∠ C = 90 degrees, square defg is connected to △ ABC, De is on AB, F and G are on BC and AC ~ B respectively In RT △ ABC, ∠ C = 90 degrees, square defg is connected to △ ABC, De is on AB, F, G are on BC and AC respectively, BC = 3, AC = 4, find ad: de: EB
- 9. In RT △ ABC, ∠ C = 90 °, AC = 3cm, BC = 4cm, then the distance from its outer center to the vertex of the right angle is______ cm．
- 10. Draw any straight line L at the right vertex C of the isosceles right triangle ABC. Make ad perpendicular to L, ad perpendicular to L, be perpendicular to L, and perpendicular to D, e ① Try to draw a figure. There are two different figures ② Explain the reason why △ ACD is equal to △ CBE respectively in the two graphs you draw
- 11. As shown in the figure, ∠ C = 90 °, B = 60 ° in △ ABC, the point O is on AB, Ao = x, and the radius of ⊙ o is 1?
- 12. If BC = 6, Ao is equal to the root sign 3, then the circle O is half
- 13. As shown in the figure, in △ ABC, ∠ C = 90 °, CA = CB, CD ⊥ AB in D, CE bisection ∠ BCD intersecting AB in E, AF bisection ∠ a intersecting CD in F. verification: EF is parallel to BC
- 14. As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons
- 15. Given that B and C are two fixed points, the absolute value of BC is equal to 8, and the circumference of triangle ABC is equal to 18, the trajectory equation of vertex A is obtained Why is there only one trajectory equation? Isn't there two cases? It is known that F1 and F2 are two focal points of the ellipse, and the point m satisfying the vector MF1 * MF2 = 0 is always inside the ellipse, then the eccentricity range of the ellipse is Why does the interior of the ellipse include boundaries?
- 16. The coordinates of the vertex B and C of the triangle ABC are (0,0) and (4,0) AB, and the length of the midline on the edge of (4,0) AB is 3
- 17. It is known that in the isosceles triangle ABC, ab = AC, ad ⊥ BC at D, CG / / AB, BG intersect ad, AC in E and f respectively, and it is proved that be 2 = EF × eg
- 18. In △ ABC, if high AD and be intersect at H point and BH = AC, then ∠ ABC=______ .
- 19. In △ ABC, AC = 10, ab = 17, BC edge height ad = 8, calculate △ ABC area (to draw a graph)
- 20. As shown in the figure, the triangle ABC is a right triangle. The area of shadow 1 is 23cm square. Find the length of BC