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In RT △ ACB, ∠ ACB = 90 °, AC = 3, BC = 4, the moving circle O passing through point C is tangent to the hypotenuse AB at the moving point P, and the value range and maximum value of CP are obtained
The distance from the point on the hypotenuse to the point C is half of the hypotenuse, 2.5 ~ 4, and the maximum value is 4
As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4, BC = 3, take a point o on BC as the center, make ⊙ o tangent to ab at e, tangent to AC at C, and the other intersection of ⊙ O and BC is D, then the length of segment BD is___ ．
If connected with OE, then: OE ⊥ AB, OE = OC ⊥ AC ⊥ OC ⊥ BeO ⊥ BCA ⊥ Boba = & nbsp; oeac ⊥ C = 90 °, AC = 4, BC = 3 ⊥ AB = 5 ⊥ bc-oeba = & nbsp; oeac ⊥ OE = 43 ⊥ OC = 43 ⊥ BD = BC-2 × OC = 13
As shown in the figure, in RT △ ABC, ACB = 90 °, BAC = 30 ° and ab = 4, point P is a moving point on the hypotenuse AB, and point D is the midpoint of CP. extend BD to e so that de = BD, connecting AE. (1) calculate the area of quadrilateral PCEA; (2) when the length of AP is what value, quadrilateral PCEA is parallelogram; (3) when the length of AP is what value, quadrilateral PCEA is right angled trapezoid
Let ch ⊥ AB be h, ∵ ACB = 90 °, ∵ BAC = 30 °, ab = 4, ∥ BC = 2, then ch = 3. Connect EP, because CD = DP, BD = De, then ▱ pbce. Then CE = Pb, EP = CB = 2. (1) sapce = (CE + AP) ch △ 2 = ab · ch △ 2 = 23, area of quadrilateral PCEA = 12 (CE + AP) · ch = 12ab · ch = 23; (2) when AP = 2, BP = EC = AP, then AP = EC, and AP ▱ EC, then ∥ PCEA, ∥ AP = 2 = PC = EC, and EC (3) when AP = 3, P and H coincide, EC ∥ AP, ∠ CPA = 90 °, AP = 3 ≠ 1 = Pb = EC, right trapezoidal PCEA is obtained; when AP = 1, △ ape is right triangle, ∠ EAP = 90 °, EC ∥ AP, AP = 1 ≠ 3 = Pb = EC, right trapezoidal PCEA is obtained
It is known that ad is the height on the hypotenuse BC of RT △ ABC, AC = 20cm, ab = 15cm. The lengths of AD, BD and CD are obtained
As shown in the figure: ∵ ABC is a right triangle, AC = 20cm, ab = 15cm, ∵ BC = AB2 + ac2 = 152 + 202 = 25cm. ∵ ad ⊥ BC, ∵ ad = ab · ACBC = 15 × 2025 = 12cm. In RT △ abd, ∵ AB = 15cm, ad = 12cm, ∵ BD = AB2 − ad2 = 152 − 122 = 9cm, ∵ BD = bc-bd = 25-9 = 16cm
In RT △ ABC, AC = 3, BC = 4, point D is a point on the hypotenuse AB, and AC = ad. (I) find the length of CD; (II) find the value of sin ∠ BDC
(1) Because in the right angle △ ABC, AC = 3, BC = 4, so AB = 5 So cosa = 35 In △ ACD, according to the cosine theorem CD2 = ac2 + ad2-2ac · adcosa (6 points) so CD2 = 32 + 32-2.3.3.35, so CD = 655 In △ BCD, SINB = 35 According to the sine theorem bcsin ∠ BDC = cdsin ∠ B (12 points) substituting BC = 4, CD = 655, we get sin ∠ BDC = 255 (13 points)
In △ ABC, ab = AC, BC = 5cm, make the middle vertical line of AB, cross the other waist AC to D, connect BD, if the circumference of △ BCD is 17cm, then the waist length is ()
A. 12cmB. 6 cmC. 7 cmD. 5 cm
As shown in the figure, the vertical bisector of ∵ AB intersects at point D, the perimeter of ∵ ad = BD, the perimeter of ∵ BCD = BD + CD + BC = AD + CD + BC = AC + BC, the perimeter of ∵ BCD is 17cm, BC = 5cm, AC = 17-5 = 12cm
As shown in the figure, ∠ ACB = 90 ° in △ ABC, ad bisects ∠ BAC, de ⊥ AB in E
It is proved that: ∵ de ⊥ AB, ∵ AED = 90 ° = ∠ ACB, and ∵ ad bisecting ∵ BAC, ∵ DAE = ∠ DAC, ∵ ad = ad, ≌ ACD, ≌ AE = AC, ∵ ad bisecting ≁ BAC, ≁ ad ⊥ CE, that is, the straight line ad is the vertical bisector of the line CE
In the triangle ABC, ∠ C is equal to 90 degrees, AC = BC, ad bisects ∠ VAB, de ⊥ AB is equal to e, and ab = 6cm,
Then the perimeter of triangle DEB is
In the triangle ABC, C is equal to 90 degrees, AC = BC, ad bisects cab, de ⊥ AB is equal to e, and ab = 6cm,
Then the perimeter of triangle DEB is, then the perimeter of triangle DEB is
Simple algorithm: because ad bisects angle a, and De is perpendicular to AB, it is easy to know that triangle ACD is congruent with triangle AED, so AC = AE, DC = De, the perimeter of triangle BDE is be + BD + de = be + BD + DC = be + BC = be + AC = be + AE = Ba = 6cm, complex algorithm ∠ C is equal to 90 degrees, AC = BC, so triangle ABC is isosceles right angle triangle AC
It is known that in the triangle ABC, ∠ BAC = 90 degrees, D is a point in the triangle ABC, and ab = AC = BD, ∠ abd = 30 degrees
Proof: make a square bace with AB and AC as sides, connect ed,
Because AB = AC = BD, angle abd = 30 degrees,
So be = AB = BD, ∠ DBE = 60 degrees,
So, triangle DBE equilateral triangle, de = BD, ∠ DEB = 60 degrees,
So, EDC = 30 degrees, CE = AB,
Therefore, the triangle abd is equal to the triangle CED, ad = DC
In △ ABC, CD ⊥ AB is D, AC = 4cm, BC = 3cm, CD = 12 / 5cm
I'll hand it in soon
Look at the picture by yourself
CD⊥AB
Delta ACD and delta abd are right triangles
Because of Pythagorean theorem
obtain
AD=√（AC²-CD²）=√（4²-12/5²）=16/5
BD=√（BC²-CD²）=√（3²-12/5²）=9/5
AB=AD+BD=16/5+9/5=5cm
RELATED INFORMATIONS
- 1. In △ ABC, ∠ C = 90 °, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time, where point P moves along AC to terminal C at the speed of 1cm / s, and point Q moves along BC to terminal C at the speed of 1.25cm/s. Passing point P, PE ‖ BC is connected to point E, and EQ is connected (1) The length of AE and De is expressed by the algebraic expression containing x; (2) when the point Q moves on BD (excluding points B and D), let the area of △ EDQ be y (cm2), find the functional relationship between Y and time x, and write out the value range of the independent variable x; (3) when the value of X is, the △ EDQ is a right angle triangle
- 2. As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
- 3. As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
- 4. As shown in the figure, the triangle ABC is an equilateral triangle, and D is the midpoint on the side ab. it is known that the area of the triangle BDE is 5 square centimeters. Find the area of the equilateral triangle ABC
- 5. In △ ABC, ab = AC, be = CF, EF intersect BC with D
- 6. As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G
- 7. As shown in the figure, in the quadrilateral ABCD, ∠ bad = ∠ ABC, ad = BC, AC intersects with point o at BD. let's say that the triangle OAB is an isosceles triangle
- 8. As shown in Figure 1, in △ ABC, points e.d.f ab.bc.ca If ad bisects ∠ BAC, then the quadrilateral AEDF is a diamond?
- 9. In the triangle ABC, ab = BC, the point D E F is the midpoint on the side of BC AC AB respectively
- 10. In the geometry shown in the figure, EA vertical plane ABC, DB vertical plane ABC, AC vertical plane BC and AC = BC = BD = 2ae. M is the midpoint of ab Finding the tangent of the angle between de and plane EMC
- 11. In a convex quadrilateral ABCD, Da = DB = DC = BC, then the degree of the largest angle in the quadrilateral is () A. 120°B. 135°C. 150°D. 165°
- 12. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 108 °, D is on AC and BC = AB + CD, the proof is: BD bisects ∠ ABC
- 13. As shown in the figure, in the isosceles trapezoid ABCD, it is known that ad ∥ BC, ab = CD, AE ⊥ BC are in E, ∠ B = 60 °, DAC = 45 ° and AC = 6?
- 14. As shown in the figure, in RT △ ABC, ∠ C = 90 °, am is the middle line on the side of BC, sin ∠ cam = 35, then the value of tanb is () A. 32B. 23C. 56D. 43
- 15. In rectangular ABCD, AE is perpendicular to BD, and the angle DAE = 2 angle BAE?
- 16. Ad is the height of triangle ABC, be is the midline angle on the side of AC, CBE = 30 degrees to prove ad = be
- 17. As shown in Figure 7-11, in the known triangle ABC, ad is perpendicular to D, BC is equal to D, AE is equal to ∠ BAC (2), if ∠ B ＞ C, try to explain ∠ DAE = 1 / 2 (∠ B - ∠ C)
- 18. It is known that in the oblique triangle ABC, the straight line with high BD and CE intersects at h, and the angle a = 45 degrees, so the degree of the angle BHC can be calculated? No picture, if you can,
- 19. In known quadrilateral ABCD, angle abd = 12 °, angle DBC = 36 °, angle ACB = 48 ° and angle ACD = 24 °, try to find the degree of angle ADB I can draw pictures according to the title,
- 20. As shown in the figure, in △ ABC, be bisects ∠ ABC intersects AC at point E, de ‖ BC intersects AB at point D, ∠ ade = 70 ° and calculates the degree of ∠ DEB Don't use the formula (one outer angle of a triangle is equal to the sum of two nonadjacent outer angles),