问题1:在ΔABC中,D和E分别是AB和AC侧的点,使得DE ||公元前。
(i)如果AD = 6厘米,DB = 9厘米,AE = 8厘米,请找到AC。
解决方案:
Given: Δ ABC where
Length of side AD = 6 cm, DB = 9 cm, AE = 8 cm
Also, DE ∥ BC,
To find : Length of side AC
By using Thales Theorem we get, (As DE ∥ BC)
AD/BD = AE/CE – equation 1
Let CE = x.
So then putting values in equation 1, we get
6/9 = 8/x
6x = 72 cm
x = 72/6 cm
x = 12 cm
∴ AC = AE + CE = 12 + 8 = 20.
Therefore, Length of side AC is 20 cm.
ii)如果AD / DB = 3/4且AC = 15 cm,则查找AE。
解决方案:
Given:
Length of side AD/BD = 3/4 and AC = 15 cm
To find: Length of side AE
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Let, AE = x
Then CE = 15 – x.
Now, putting values in equation 1,
⇒ 3/4 = x/ (15 – x)
45 – 3x = 4x
-3x – 4x = – 45
7x = 45
x = 45/7
x = 6.43 cm
∴ AE= 6.43cm
Therefore, Length of side AE is 6.43 cm
iii)如果AD / DB = 2/3且AC = 18 cm,则查找AE。
解决方案:
Given:
Length of side AD/BD = 2/3 and AC = 18 cm
To find: Length of side AE.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Let, AE = x
Then CE = 18 – x
Now, putting values in equation 1,
⇒ 23 = x/ (18 – x)
3x = 36 – 2x
5x = 36 cm
x = 36/5 cm
x = 7.2 cm
∴ AE = 7.2 cm
Therefore, Length of side AE is 7.2 cm
iv)如果AD = 4厘米,AE = 8厘米,DB = x – 4厘米,EC = 3x – 19,找到x。
解决方案:
Given:
Length of side AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19
To Find: length of side x.
By using Thales Theorem, we get,
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
4/ (x – 4) = 8/ (3x – 19)
4(3x – 19) = 8(x – 4)
12x – 76 = 8(x – 4)
12x – 8x = – 32 + 76
4x = 44 cm
x = 11 cm
Therefore, Length of side x is 11 cm
v)如果AD = 8厘米,AB = 12厘米和AE = 12厘米,请找到CE。
解决方案:
Given:
Length of side AD = 8 cm, AB = 12 cm, and AE = 12 cm.
To find: Length of side CE.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
8/4 = 12/CE
8 x CE = 4 × 12 cm
CE = (4 × 12)/8 cm
CE = 48/8 cm
∴ CE = 6 cm
Therefore, Length of side x is 6 cm
vi)如果AD = 4厘米,DB = 4.5厘米和AE = 8厘米,请找到AC。
解决方案:
Given:
Length of side AD = 4 cm, DB = 4.5 cm, AE = 8 cm
To find: Length of side AC.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
4/4.5 = 8/AC
AC = (4.5 × 8)/4 cm
∴AC = 9 cm
Therefore, Length of side AC is 9 cm
vii)如果AD = 2厘米,AB = 6厘米,AC = 9厘米,则找到AE。
解决方案:
Given:
Length of side AD = 2 cm, AB = 6 cm and AC = 9 cm
To find: Length of side AC.
Length of DB = AB – AD = 6 – 2 = 4 cm
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
2/4 = x/(9 – x)
4x = 18 – 2x
6x = 18
x = 3 cm
∴ AE = 3cm
Therefore, Length of side AE is 3 cm
viii)如果AD / BD = 4/5且EC = 2.5厘米,则查找AE。
解决方案:
Given:
Length of side AD/BD = 4/5 and EC = 2.5 cm
To find: Length of side AC.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
Then, 4/5 = AE/2.5
∴ AE = 4 × 2.55 = 2 cm
Therefore, Length of side AE is 2 cm
ix)如果AD = x cm,DB = x – 2 cm,AE = x + 2 cm,EC = x – 1 cm,则求出x的值。
解决方案:
Given:
Length of side AD = x, DB = x – 2, AE = x + 2 and EC = x – 1
To find: Value of x.
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
x/(x – 2) = (x + 2)/(x – 1)
x(x – 1) = (x – 2)(x + 2)
x 2– x – x2 + 4 = 0
∴ x = 4
Therefore, the value of x is 4
x)如果AD = 8x – 7厘米,DB = 5x – 3厘米,AE = 4x – 3厘米,EC =(3x – 1)厘米,则求出x的值。
解决方案:
Given:
Length of side AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1
To find : Value of x
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
(8x – 7)/(5x – 3) = (4x–3)/ (3x–1)
(8x – 7)(3x – 1) = (5x – 3)(4x – 3)
24x2 – 29x + 7 = 20x2– 27x + 9
4x2 – 2x – 2 = 0
2(2x2 – x – 1) = 0
2x2 – x – 1 = 0
2x2 – 2x + x – 1 = 0
2x(x – 1) + 1(x – 1) = 0
(x – 1)(2x + 1) = 0
⇒ x = 1 or x = -1/2
Since, we know that the side of triangle is always positive.
Therefore, we take the positive value.
∴ x = 1.
Therefore, the value of x is 1.
xi)如果AD = 4x – 3,AE = 8x – 7,BD = 3x – 1,CE = 5x – 3,则求出x的值。
解决方案:
Given:
Length of side AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3
To find : Value of x
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
We get,
(4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)
(4x – 3)(5x – 3) = (3x – 1)(8x – 7)
4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7)
20x2 – 12x – 15x + 9 = 24x2 – 29x + 7
20x2 -27x + 9 = 24x2 – 29x + 7
⇒ -4x2+ 2x + 2 = 0
4x2 – 2x – 2 = 0
4x2 – 4x + 2x – 2 = 0
4x(x – 1) + 2(x – 1) = 0
(4x + 2)(x – 1) = 0
⇒ x = 1 or x = -2/4
We know that the side of triangle is always positive
Therefore, we only take the positive value.
∴ x = 1
Therefore, the value of x is 1.
xii)如果AD = 2.5厘米,BD = 3.0厘米和AE = 3.75厘米,请找到AC的长度。
解决方案:
Given:
Length of side AD = 2.5 cm, AE = 3.75 cm and BD = 3 cm
To find : Length of side AC
By using Thales Theorem, we get
AD/BD = AE/CE – equation 1
Now, putting values in equation 1,
We get,
2.5/ 3 = 3.75/ CE
2.5 x CE = 3.75 x 3
CE = 3.75×32.5
CE = 11.252.5
CE = 4.5
Now, AC = 3.75 + 4.5
∴ AC = 8.25 cm.
Therefore, the value of AC is 8.25cm
问题2.在ΔABC中,D和E分别是边AB和AC上的点。对于以下每种情况,请显示DE∥BC:
i)AB = 12厘米,AD = 8厘米,AE = 12厘米,AC = 18厘米。
解决方案:
Given:
Length of side AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm
To show : Side DE is parallel to BC ( DE || BC )
Now,
Length of side BD = AB – AD
⇒ 12 – 8 = 4 cm
And,
Length of side CE = AC – AE = 18 – 12 = 6 cm
Now,
AD/BD = 8/4 = 1/2 – equation 1
AE/CE = 12/6 = 1/2 – equation 2
Now from equation 1 and 2
AD/BD = AE/CE
So, by the converse of Thale’s Theorem
We get, DE ∥ BC.
Hence, Proved.
ii)AB = 5.6 cm,AD = 1.4 cm,AC = 7.2 cm,AE = 1.8 cm。
解决方案:
Given:
Length of side AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm
To show : Side DE is parallel to BC ( DE || BC )
Now,
Length of side BD = AB – AD = 5.6 – 1.4 = 4.2 cm
And,
Length of side CE = AC – AE = 7.2 – 1.8 = 5.4 cm
Now,
AD/BD = 1.4/4.2 = 1/3 – equation 1
AE/CE = 1.8/5.4 =1/3 – equation 2
Now from equation 1 and 2
AD/BD = AE/CE
So, by the converse of Thale’s Theorem
We get, DE ∥ BC.
Hence, Proved.
iii)AB = 10.8厘米,BD = 4.5厘米,AC = 4.8厘米,AE = 2.8厘米。
解决方案:
Given:
Length of side AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.
To show: Side DE is parallel to BC ( DE || BC )
Now,
Length of side AD = AB – DB = 10.8 – 4.5 = 6.3
And,
Length of side CE = AC – AE = 4.8 – 2.8 = 2
Now,
AD/BD = 6.3/ 4.5 = 2.8/ 2.0 = 7/5 – equation 1
AE/CE = 7/5 – equation 2
Now from equation 1 and 2
AD/BD = AE/CE
So, by the converse of Thale’s Theorem
We get, DE ∥ BC.
Hence Proved
iv)AD = 5.7cm,BD = 9.5cm,AE = 3.3cm,EC = 5.5cm。
解决方案:
Given:
Length of side AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm
To show: Side DE is parallel to BC (DE || BC)
Now,
Length of side AD/BD = 5.7/9.5 = 3/5 – equation 1
And,
Length of side AE/CE = 3.3/5.5 = 3/5 – equation 2
Now from equation 1 and 2
AD/BD = AE/CE
So, by the converse of Thale’s Theorem
We get, DE ∥ BC.
Hence, Proved
问题3:在ΔABC中,P和Q分别是AB和AC侧的点,因此PQ∥BC。如果AP = 2.4厘米,AQ = 2厘米,QC = 3厘米,BC = 6厘米。查找AB和PQ。
解决方案:
Given:
In ΔABC,
Length of side AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.
Also, PQ ∥ BC.
To find: Length of side AB and PQ
Now,
Since it’s given that PQ ∥ BC
So by using Thales Theorem, we get
AP/PB = AQ/ QC
2.4/PB = 2/3
2 x PB = 2.4 × 3
PB = (2.4 × 3)/2 cm
⇒ PB = 3.6 cm
Therefore, Length of PB is 3.6 cm
Now finding, AB = AP + PB
AB = 2.4 + 3.6
⇒ AB = 6 cm
Therefore, Length of AB is 6 cm
Now, considering ΔAPQ and ΔABC
We have,
∠A = ∠A (Common angle)
∠APQ = ∠ABC (Corresponding angles are equal and PQ||BC and AB being a transversal)
Thus, ΔAPQ and ΔABC are similar to each other by AA criteria.
Now, we know that corresponding parts of similar triangles are propositional.
Therefore,
⇒ AP/AB = PQ/ BC
⇒ PQ = (AP/AB) x BC
= (2.4/6) x 6 = 2.4
∴ PQ = 2.4 cm.
Therefore, Length of PQ is 2.4 cm and AB is 6cm
问题4:在ΔABC中,D和E分别是AB和AC上的点,因此DE∥BC。如果AD = 2.4厘米,AE = 3.2厘米,DE = 2厘米和BC = 5厘米。查找BD和CE。
解决方案:
Given:
In Δ ABC,
Length of side AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm
Also, DE ∥ BC
To find: Length of side BD and CE.
As DE ∥ BC, AB is transversal,
∠APQ = ∠ABC (corresponding angles) – equation 1
As DE ∥ BC, AC is transversal,
∠AED = ∠ACB (corresponding angles) – equation 2
In Δ ADE and Δ ABC,
Now from equation 1 and 2 we get,
∠ADE = ∠ABC
∠AED = ∠ACB
∴ ΔADE = ΔABC (By AA similarity criteria)
Now, we know that
Corresponding parts of similar triangles are proportional.
Therefore,
⇒ AD/AB = AE/AC = DE/BC
AD/AB = DE/BC
2.4/ (2.4 + DB) = 2/5 [Since, length of side AB = AD + DB]
2.4 + DB = 6
DB = 6 – 2.4
DB = 3.6 cm
Length of side DB is 3.6 cm
In the same way, we get
⇒ AE/AC = DE/BC
3.2/ (3.2 + EC) = 2/5 [Since AC = AE + EC]
3.2 + EC = 8
EC = 8 – 3.2
EC = 4.8 cm
∴ BD = 3.6 cm and CE = 4.8 cm.
Length of side BD is 3.6 cm and CE is 4.8 cm
问题5:对于给定的数字,请陈述PQ∥EF。
解决方案:
Given :
Length of side EP = 3 cm, PG = 3.9 cm, FQ = 3.6 cm and QG = 2.4 cm
To find: PQ ∥ EF or not.
According to Thales Theorem,
PG/GE = GQ/FQ
Now,
3.9/3 ≠ 3.6/ 2.4
As we can see it is not proportional.
So, PQ is not parallel to EF.
问题6:M和N分别是ΔPQR边PQ和PR上的点。对于以下每种情况,请声明MN∥QR。
(i)PM = 4厘米,QM = 4.5厘米,PN = 4厘米,NR = 4.5厘米。
(ii)PQ = 1.28 cm,PR = 2.56 cm,PM = 0.16 cm,PN = 0.32 cm。
(i)
Given:
In the ∆PQR
M and N are points on PQ and PR respectively
PM = 4 cm, QM = 4.5 cm, PN = 4 cm, RN = 4.5 cm
To find: MN || QR or not
According to Thales Theorem,
PM/QM = PN/NR
4/4.5 = 4/4.5
Hence, MN || QR
(ii)
Given :
Length of side PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, and PN = 0.32 cm.
To find: that MN ∥ QR or not.
MQ = PQ – PM = 1.28 – 0.16 = 1.12 cm
RN = PR – PN = 2.56 – 0.32 =2.24 cm
According to Thales Theorem,
PM/QM = PN/ RN
PM/QN = 0.16/1.12 = 1/7 – equation 1
PN/RN = 0.32/ 2.24 = 1/7 – equation 2
From equation 1 and 2
PM/QM = PN/ RN
Therefore, MN || QR
问题7:在三线段OA,OB和OC中,分别选择点L,M,N,使得LM∥AB和MN BC,但L,M和N以及A,B,C都不共线。显示LN∥AC。
解决方案:
Given :
OA, OB and OC, points are L, M, and N respectively
Such that LM || AB and MN || BC
To prove: LN ∥ AC
Now,
In ΔOAB, Since, LM ∥ AB,
Then, OL/LA = OM/ MB (By using BPT) – equation 1
In ΔOBC, Since, MN ∥ BC
Then, OM/MB = ON/NC (By using BPT)
Therefore, ON/NC = OM/ MB – equation 2
From equation 1 and 2, we get
OL/LA = ON/NC
Therefore, In ΔOCA By converse BPT, we get
LN || AC
Hence proved
问题8:如果D和E分别是ΔABC的AB和AC侧的点,则DE∥BC和BD = CE。证明ΔABC是等腰的。
解决方案:
Given :
In Δ ABC, DE ∥ BC and BD = CE.
To prove: that Δ ABC is isosceles.
According to Thales Theorem,
AD/DB = AE/EC
But BD = CE (Given)
Therefore, we get,AD = AE
Now, BD = CE and AD = AE.
So, AD + BD = AE + CE.
Therefore, AB = AC.
Therefore, ΔABC is isosceles.
Hence proved