2 It is known that the three sides of △ABC satisfy the relation A? +b？ c？ -a？ c？ -B4 (the fourth power of B) =0, try to judge the shape of △ABC?
A bamboo is ten feet high. A gust of wind broke the bamboo, and its tip just hit the ground, 3 feet away from the original bamboo. How tall was the bamboo in the original place?
4 If the ratio of the degrees of the three external angles of triangle ABC is 3∶4∶5, the relationship between the largest side AB and the smallest side BC is
5 is the known equilateral triangle ABC. AB=6. Find the area of this triangle and the distance from the midpoint on one side to the other.
6 in the triangle
At ABC. ACB angle =90 degrees. CD is perpendicular to point d, if AC= 16. BC= 12. Find the length of CD.
It is known that AC and BD are perpendicular to each other and connected with points O and AB. BC.CD.DA，
Verification: Party AB+Party CD = Party BC+Party AD.
8 It is known that at RT△ABC, ∠C=90 degrees, M is the midpoint of BC, and if it passes M, MD⊥AB is at D.
Please explain the three-line advertisement. BD and AC can always form a right triangle.
9 As shown in the figure, it is known that in △ABC, ∠ c = 90, ∠ 1=∠2, CD= 1.5, BD=2.5, and find the length of AC.
10 As shown in the figure, in the Rt triangle ABC, the angle C=90 degrees, D is the midpoint of AB, E and F are on AC and BC respectively, and DE is perpendicular to D F.. Try to judge the relationship between AE, EF and FB and prove it.
answer
1
∫∠DCP 90，DC=CP
∴dp=√(dc^2+dp^2)=√(2^2+2^2)=2√(2)
∴∠CDP=∠CPD=45
∠∠DCP =∠ACB = 90
∴∠DCP-∠PCB=∠ACB-∠PCB
That is ∠ACP=∠BCD.
CD = CP，AB=BC。
∴△ACP≌△BDC
∴PA=BD=3
∵(2√(2))^2+ 1^2=3^2
∴DP^2+BP^2=DB^2
∴∠DPB=90
∠∠BPC =∠CPD+∠DPB
∴∠BPC=45 +90 = 135
2
a^2+b^2c^2-a^2c^2-b^4=a^2( 1-c^2)+b^2(c^2-b^2)=0
∫a2, B 2 and C 2 are not equal to zero.
∴ 1-c^2=0
c^2-b^2=0
∴b^2=c^2= 1
△ABC is an isosceles triangle
three
Solution: The original place is x feet tall with bamboo.
x^2+3^2=( 10-x)^2
solve
X=4.55 feet
A: There are 4.55-foot-tall bamboos in the same place.
four
The degree of the outer angle of a triangle is equal to the sum of the other two inner angles, so the sum of the three outer angles should be equal to 360. Given that the ratio of degrees of the three external angles is 3∶4∶5, we can deduce that the degrees of these three angles are 90, 120 and 150 respectively. That is, the three internal angles of this triangle are 30, 60 and 90 degrees respectively.
So the relationship between the largest side AB and the smallest side BC is AB=2BC.
five
Area 9 times root number 3
The distance is longer than the root mulberry 1.5 times.
Process: All three sides are 6, the angle is 60 degrees, and the square of height = 6 square -3 square.
3 times higher than the root number 3.
Area = edge times height divided by 2=9 times root number 3
Because each angle is 60 degrees, the sides of a 30-degree right triangle pair are half of the hypotenuse, so the square of the distance from the midpoint of one side to the other side is equal to the square of 3-1.5.
The distance from the midpoint on one side to the other side is long = 1.5 times the root number 3.
six
∫AC = 16
BC= 12
triangle
The ABC area is 192.
The other angle ACB=90 degrees
The length of AB can be found to be 20 by Pythagorean theorem.
AB*CD=AC*BC can be obtained from the area formula.
The ∴CDC length is 9.6.
seven
Because of ACBD,
So ab 2 = bo 2+ao 2, CD 2 = co 2+do 2,
So ab 2+CD 2 = bo 2+ao 2+co 2+do 2 = (bo 2+co 2)+(ao 2+do 2) = BC 2+ad 2.
eight
Connect am, ∠ c = 90 degrees,
Because ∠C=90 degrees,
So AC 2+cm 2 = am 2,
Because m is the midpoint of BC,
So BM=CM,
So AC 2+BM 2 = am 2,
Because MD⊥AB,
So BM 2 = MD 2+BD 2,
So AC 2 = AM 2-MC 2 = AM 2-MB 2 = AM 2-(MD 2+BD 2) = AM 2-MD 2-BD 2 = AD 2-BD 2,
So AC 2+BD 2 = AD 2,
So ad. BD and AC can always form a right triangle.
nine
Let DE vertical AB, AAS and AED congruent ACD, DE= 1.5, DB=2.5, Pythagoras EB=2, tanB=0.75.
CB=4，AC=3
10
Ideas:
FD is extended to k, so that DK=DF, even EK, AK,
Triangle EFD congruent triangles EKD,
EF=EK，
Triangle BDF congruent triangles ADK,
AK=BF，
Angle A+ angle B = 90
So the triangle AEK is a right triangle.
AE^2+AK^2=EK^2，
AE^2+BF^2=EF^2
Landlord! Very detailed! O(∩_∩)O Haha ~, but the first, ninth and tenth questions have pictures, so you can't send them. . . . The landlord gave me a mailbox and I sent it to the landlord, okay? O(∩_∩)O~