😊 Amazing problem! I really like this! This was a bit challenging! I am happy that here we can use the secant - secant theorem! Hope to get new problems everyday where we can use different theorems. All the best!
@soli9mana-soli4953 Two triangles are similar If you can demonstrate, at least, that two of its angles are equals. ∡CDB = ½ ∡COB ∡CBE = ½ ∡COE and ∡COB = ∡COE
As long as chords BC and CE are equal in length, any locations of B and C are valid. The problem statement implies that x is the same for all valid locations. As B and C become closer and closer to each other, the limiting case is where they merge into the same point. In that case, AB is a tangent and we apply the tangent-secant theorem to find that the tangent length is 6. ABC is a right triangle. Applying the Pythagorean theorem, AC has length 9, AB length 6 and BC length x., so 9² = 6² + x², 81 = 36 + x², x² = 81 - 36 = 45, x = √(45) = 3√5, as PreMath also found. On a multiple choice test, choose 3√5 and go on. If the test does not require you to show your work, just provide x = 3√5 as an answer. If you must show your work, next solve the general case. This solution method may be helpful for finding the solution to the general case. It also serves as a check that you solved the problem correctly.
**Enjoyed sol offered. Now like to forward a sol using Theorem on similar Triangles ** We may join BD Then we may get two similar triangles ABD & ACE [ as ang BAC is common, ang DBE =ang DCE as these are inscribed angle on the same arc DE ] Then 4/t =( 2m+t)/(4+5) >4/t =(2m+t)/9 > 2mt +t^2 =36 -- (1) In rt triangle ACF hypotenuse AC^2 =CF^2+(AE + EF) ^2 =[x ^2 - m^2+(t +m) ^2] =x ^2-m^2+t^2+2mt+m^2 =x ^2 +t^2+2mt =x ^2 +36 (pl see eq 1) Hence x ^2 +36=81[as AC^2=(4+5)^2] > x=√45=3√5 units
setting AE = a , EB = b and angles in B and C = alpha, I solved in this way: applying secant-secant theorem we have: 9*4 = (a+b)*a 36 = a² + ab applying sines law on BCE: x/sin alpha = b/sin(180-2alpha) and knowing that sin(180-2alpha)=sin 2alpha and sin 2 alpha = 2 sin alpha cos alpha we get cos alpha = b/2x applying cosine rule on ABC we have: (4+5)² = x² + (a+b)² - 2x*(a+b)*b/2x 81 = x² + a² + ab (that was a lucky result I didn't count on at the beginning, being started with 3 unknowns!) 81 = x² + 36 x = 3sqrt5
Can you help to solve this difficult probability question. In a bag, there were 6 red, 8 yellow and 10 blue balls. one after another, balls are removed without replacement. Find (a) the probability that red is completely removed before yellow; and (b) the probability that red is completely removed before yellow and blue. Thanks.
Let's find x: . .. ... .... ..... According to the intersecting secants theorem we know: AE*AB = AD*AC AE*(AE + BE) = AD*(AD + CD) = 4*(4 + 5) = 4*9 = 36 Let M be the midpoint of BE. Since BCE is an isosceles triangle, we know that BCM and CEM are congruent right triangles. So we can conclude: BM = EM ⇒ BE = BM + EM = 2*EM Since ∠AMC=∠EMC=90°, ACM is also a right triangle. By applying the Pythagorean theorem to the right triangles ACM and CEM we obtain: CM² + AM² = AC² CM² + (AE + EM)² = (AD + CD)² CM² + AE² + 2*AE*EM + EM² = (4 + 5)² CM² + AE² + AE*2*EM + EM² = 9² CM² + AE² + AE*BE + EM² = 81 CM² + AE*(AE + BE) + EM² = 81 CM² + 36 + EM² = 81 CM² + EM² = 45 x² = CE² = CM² + EM² = 45 ⇒ x = √45 = 3√5 Best regards from Germany
Solution: We know that BC is equal to CE, thus BCE is an isosceles triangle. From C we draw a perpendicular until the point F on the line AB, and we label the length as h, such that F is EB midpoint. EF = k BF = k Applying Pythagorean Theorem in Triangle BCE, we have: k² + h² = x² ... ¹ Let's label AE = z Applying, from A, Secant Theorem, we have: AD × AC = AE × AB 4 × (4 + 5) = z × (z + k + k) 4 × (9) = z × (z +2k) z² + 2zk = 36 ... ² Applying Pythagorean Theorem once again in Triangle ACF, we have: (AE + EF)² + (CF)² = (AD + CD)² (z + k)² + (h)² = (4 + 5)² z² + 2zk + k² + h² = 9² (z² + 2zk ... replacing from Eq ² = 36) + (k² + h² ... replacing from Eq ¹ = x²) = 9² 36 + x² = 81 x² = 45 x = 3√5 Units ✅ x ≈ 6,7082 Units ✅
Hello, I was trying to find the radius of the circle. What I got as the result was sqrt(45/2), because F has to be the intersection point of the perpendicular through the mid point between D and C, and the Thales circle above AC with radius 9/2. So the conclusion would be, that the center point O of the circle and point F are congruent. If I'm not wrong? Did anyone else check? Best greetings and wishes!
Appreciate the introduction to the Secant-Secant Theorem.
Sure!
Glad to hear that!
Thanks for the feedback ❤️
😊 Amazing problem! I really like this! This was a bit challenging! I am happy that here we can use the secant - secant theorem! Hope to get new problems everyday where we can use different theorems. All the best!
△ABC is similar to △BCD
1) ∡ACB is common, ∡ACB = ∡DCB
2) ∡CDB = ∡ABC, because ∡COB = ∡COE (isosceles triangle BCE)
Therefore::
x/9 = 5/x
x² = 45 --> x= 3√5 cm (Solved √)
Great! 👍 even if I didn’t caught your explanation about the similarity
@soli9mana-soli4953
Two triangles are similar If you can demonstrate, at least, that two of its angles are equals.
∡CDB = ½ ∡COB
∡CBE = ½ ∡COE
and ∡COB = ∡COE
@ thank you for the explanation, your solution is nice 👍
Looks like a proof of the secant theorem?
Thanks for the introduction of secant secant theorem
Similarity of triangles:
x/9 = 5/x
x² = 45 --> x = 3√5 cm ( Solved √ )
△ABC is similar to △BCD
1) ∡ACB is common, ∡ACB = ∡DCB
2) ∡CDB = ∡ABC, because ∡COB = ∡COE (isosceles triangle BCE)
Too complicated video solution.
Nice problem sir ❤
As long as chords BC and CE are equal in length, any locations of B and C are valid. The problem statement implies that x is the same for all valid locations. As B and C become closer and closer to each other, the limiting case is where they merge into the same point. In that case, AB is a tangent and we apply the tangent-secant theorem to find that the tangent length is 6. ABC is a right triangle. Applying the Pythagorean theorem, AC has length 9, AB length 6 and BC length x., so 9² = 6² + x², 81 = 36 + x², x² = 81 - 36 = 45, x = √(45) = 3√5, as PreMath also found.
On a multiple choice test, choose 3√5 and go on. If the test does not require you to show your work, just provide x = 3√5 as an answer. If you must show your work, next solve the general case. This solution method may be helpful for finding the solution to the general case. It also serves as a check that you solved the problem correctly.
@jimlocke9320: thank you so much!
Shaandaar,
Excellent.
Excellent.
If you drop a perpendicular from o to EB , it bisects EB so CF passes through o
**Enjoyed sol offered. Now like to forward a sol using
Theorem on similar Triangles **
We may join BD
Then we may get two similar triangles ABD & ACE [ as ang BAC is common, ang DBE =ang DCE as these are inscribed angle on the same arc DE ]
Then
4/t =( 2m+t)/(4+5)
>4/t =(2m+t)/9
> 2mt +t^2 =36 -- (1)
In rt triangle ACF hypotenuse
AC^2 =CF^2+(AE + EF) ^2
=[x ^2 - m^2+(t +m) ^2]
=x ^2-m^2+t^2+2mt+m^2
=x ^2 +t^2+2mt
=x ^2 +36 (pl see eq 1)
Hence x ^2 +36=81[as AC^2=(4+5)^2]
> x=√45=3√5 units
Similarity of triangles:
x/9 = 5/x
x² = 45 --> x = 3√5 cm
△ABC is similar to △BCD
1) ∡ACB is common, ∡ACB = ∡DCB
2) ∡CDB = ∡ABC, because ∡COB = ∡COE (isosceles triangle BCE)
Therefore:: x/9 = 5/x
@@marioalb9726 will see and then will respond. I think we chatted on another occasion isn"t?
@@marioalb9726 OK, may offer this solution as an another method in the main format.
I enjoyed it.
The altitude that you dropped from C down to EB must go through the center of the circle, because of symmetry.
setting AE = a , EB = b and angles in B and C = alpha, I solved in this way:
applying secant-secant theorem we have:
9*4 = (a+b)*a
36 = a² + ab
applying sines law on BCE:
x/sin alpha = b/sin(180-2alpha)
and knowing that sin(180-2alpha)=sin 2alpha
and sin 2 alpha = 2 sin alpha cos alpha
we get
cos alpha = b/2x
applying cosine rule on ABC we have:
(4+5)² = x² + (a+b)² - 2x*(a+b)*b/2x
81 = x² + a² + ab (that was a lucky result I didn't count on at the beginning, being started with 3 unknowns!)
81 = x² + 36
x = 3sqrt5
Linda questão
Bom dia Mestre
Show de exercício
Triangles CDB and CBA are similar so CD/CB=CB/CA=DB/DA so CB*CB=CA*CD=45
Can you help to solve this difficult probability question.
In a bag, there were 6 red, 8 yellow and 10 blue balls.
one after another, balls are removed without replacement.
Find (a) the probability that red is completely removed before yellow; and (b) the probability that red is completely removed before yellow and blue. Thanks.
Let's find x:
.
..
...
....
.....
According to the intersecting secants theorem we know:
AE*AB = AD*AC
AE*(AE + BE) = AD*(AD + CD) = 4*(4 + 5) = 4*9 = 36
Let M be the midpoint of BE. Since BCE is an isosceles triangle, we know that BCM and CEM are congruent right triangles. So we can conclude:
BM = EM ⇒ BE = BM + EM = 2*EM
Since ∠AMC=∠EMC=90°, ACM is also a right triangle. By applying the Pythagorean theorem to the right triangles ACM and CEM we obtain:
CM² + AM² = AC²
CM² + (AE + EM)² = (AD + CD)²
CM² + AE² + 2*AE*EM + EM² = (4 + 5)²
CM² + AE² + AE*2*EM + EM² = 9²
CM² + AE² + AE*BE + EM² = 81
CM² + AE*(AE + BE) + EM² = 81
CM² + 36 + EM² = 81
CM² + EM² = 45
x² = CE² = CM² + EM² = 45 ⇒ x = √45 = 3√5
Best regards from Germany
شكرا لكم على المجهودات
المستقيم (CF) مار من النقطة O.
Much fun. Makes me feel like a clever detective solving a complex whodunit.
Please proof the secant theorem .
Solution:
We know that BC is equal to CE, thus BCE is an isosceles triangle. From C we draw a perpendicular until the point F on the line AB, and we label the length as h, such that F is EB midpoint.
EF = k
BF = k
Applying Pythagorean Theorem in Triangle BCE, we have:
k² + h² = x² ... ¹
Let's label AE = z
Applying, from A, Secant Theorem, we have:
AD × AC = AE × AB
4 × (4 + 5) = z × (z + k + k)
4 × (9) = z × (z +2k)
z² + 2zk = 36 ... ²
Applying Pythagorean Theorem once again in Triangle ACF, we have:
(AE + EF)² + (CF)² = (AD + CD)²
(z + k)² + (h)² = (4 + 5)²
z² + 2zk + k² + h² = 9²
(z² + 2zk ... replacing from Eq ² = 36) + (k² + h² ... replacing from Eq ¹ = x²) = 9²
36 + x² = 81
x² = 45
x = 3√5 Units ✅
x ≈ 6,7082 Units ✅
Hello,
I was trying to find the radius of the circle.
What I got as the result was sqrt(45/2), because F has to be the intersection point of the perpendicular through the mid point between D and C, and the Thales circle above AC with radius 9/2.
So the conclusion would be, that the center point O of the circle and point F are congruent.
If I'm not wrong?
Did anyone else check?
Best greetings and wishes!
John a raison o appartient à la médiatrice du triangle isocèle inscrit dans le cercle