A Very Nice Geometry Problem | You should be able to solve this | 2 Methods

A Very Nice Geometry Problem | You should be able to solve this | 2 Methods
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[@Mediterranean81]

Use the tangent secant theorem
x is the tangent
BD is the secant
So AD=x^2/4
AB= 4+x^2/4
AB^2= 16+2x^2+x^4/16 (1)
ABC is a right triangle
So AB^2=BC^2+AC^2
AB^2 = 36+x^2 (2)
Let x^2= u
Compare (1) and (2)
16+2u+u/16=36+u
256+32u+u=16u+576
32u+u-16u=576-256
15u=320
3u=64
u=64/3
x=8/sqrt 3

[@WahranRai]

9:11 The theorem is the *Power of a point* relative to a circle

[@brettgbarnes]

Based on the starting diagram and information, it cannot be deduced that AC is tangent to the circle at point C. That assumption was not given until after the solution began.

[@fphenix]

I started like Method 1, but once I had the 2√5 length, I used the tan() in B.=> tan(DBC) = 2√5 / 4 = √5/2 = tan(ABC) = x / 6. Hence 2.x = 6.√5, so x = 3√5

[@Captain_Thunder_22]

Wow ... Good video ❤

[@marcgriselhubert3915]

O the center of the circle, and t = angleCBA,
In triangle OBD: OD^2 = BO^2 + BD^2 -2.BO.BD.cos(t)
So, 9 = 9 + 16 - 2.3.4.cos(t), and cos(t) = 2/3
Then 1 + (tan(t))^2 = 1/(cos(t))^2 = 9/4,
then (tan(t))^2 = 5/5 and tan(t) = sqrt(5)/2
In triangle CBA: X = CA = BC.tan(t) = 6.(sqrt(5)/2)
Finally: X = 3.sqrt(5).

[@bpark10001]

In 1st method, you need not solve for DC. It is simpler to solve for AB. Using same triangle similarity, BD/BC = BC/AD. From this you get BD = 9. Then you can use Pythagoras to get AC squared = 81 - 36 = 45.

[@rabotaakk-nw9nm]

👍, but BD/BC=BC/AB => AB=9
Or, AD=AB-BD=9-4=5
AC²=AD•AB=5•9=45 😉

[@machintruc8302]

Cos(dbc) = cos(abc) = BD/BC = BC/BA
So BA=BC²/BD=6²/4=9
Then
AB²=BC²+AC²
9²=6²+x²
x²=81-36=45
x=v(45)=3v(5)

[@murdock5537]

∆ ABC → BC = 6; AB = AD + BD = y + 4; CD = z; AC = x; ABC = δ
sin⁡(BCA) = sin⁡(CDB) = 1 → z = 2√5 → tan⁡(δ) = z/4 = √5/2 = x/6 → x = 3√5 → y = 5

[@santiagoarosam430]

Llamamos E a la proyección ortogonal de D sobre BC → 6²-4²=DC²→ DC=2√5 → BC*DE=BD*DC→ DE=4√5/3 → BD²-DE²=BE²→ BE=8/3 → DE/BE=AC/BC→AC=X=3√5 =6,7082....
Gracias y un saludo cordial.

[@user-ll5wl9gq4x]

Thanks for the beautiful ideas!!

[@imetroangola4943]

Parabéns 💐👏🏻 são excelentes seus vídeos!

[@JobBouwman]

Thales tells DC = sqrt(6^2 - 4^2)=2sqrt5.
Congruency tells that AC = 6/4*2sqrt5.
So AC = 3sqrt(5)

[@quigonkenny]

Let O be the center of the semicircle. Draw radius OD. OD = OB = BC/2 = 6/2 = 3. As OB = OD, ∆BOD is an isosceles triangle and ∠DBO = ∠ODB = θ.
Triangle ∆BOD:
cos(θ) = (OB²+DB²-OD²)/(2(OB)DB)
cos(θ) = (3²+4²-3²)/(2(3)4)
cos(θ) = (9+16-9)/24
cos(θ) = = 16/24 = 2/3
sin(θ) = √(1-cos²(θ))
sin(θ) = √(1-(2/3)²)
sin(θ) = √(1-4/9)
sin(θ) = √(5/9) = √5/3
tan(θ) = sin(θ)/cos(θ)
tan(θ) = (√5/3)/(2/3) = √5/2
CA/BC = √5/2
√5BC = 2CA
6√5 = 2x
x = 6√5/2 = 3√5 units

[@giuseppemalaguti435]

DC=√20...4:√20=√20:AD..AD=5..x^2=(4+5)^2-6^2=45

[@sergeyvinns931]

Нарисовано коряво! Но DC^2=36-16=20, DC=2\/5, DC/x=4/6, 2\/5/х=2/3, 2х=6\/5, х=3\/5, найдём гипотенузу АВ,
AB^2=[^2+BC^2=45+36=81, AB=9. А нарисовано так, что АС=ВС?

[@yakupbuyankara5903]

X=3×(5^(1/2)).

[@haiduy7627]

🎉🎉🎉🎉🎉🎉🎉

[@Tmwyl]

I got this one!

[@michaeldoerr5810]

Based on one of the liked comments from Math Booster, there might actually be THREE methods. Just like the last video. I shall have to practice all three methods then!

[@haiduy7627]

❤❤❤❤❤❤❤🎉🎉😊😊😊❤❤❤❤🎉🎉🎉🎉

[@RealQinnMalloryu4]

(4)^2H/A/BDASino°=16H/A/DBASino° (6)^2 A/H/BCoso°= 36A/H/BCCoso° {16H/A/BDASino°+36A/H/BCCoso°}= 52HA/AH/BDASino°BCCoso° {180°/52HA/AH/BDASino°BCCoso°} =3 .24 HA/AH/BDASino°BCCoso° 3 .4^6 3.2^23^2: 1.1^1 3^2 3^2: (HA/AHSino°BDABCCoso° ➖ 3HA/AH/BDASino°BCCoso°+2)