4/AC= 2/DC(theorem). Therefore AC= 2 DC. AD= sqroot 20(pythag.). Then AD^2= AB.AC- BD.DC so 20= 4.2DC- 2DC= 6DC.Then DC= 10/3. So BC=16/3. Area of triangleABC = (1/2) by 4 by 16/3= 32/3.

Hello. Many thanks for the video. Excellent work as usual. The 3rd method can be hugely simplified. The triangle DEC being similar to ABC with a factor of 2/4=1/2, it has a area of 1/4 of A, Area of ABC. And so, A = 4 + 4 + A/4 = 32/3

Thank you professor.

Congratiulations!! The 3º method, using auxiliary lines is terrific!!!

*I like to solve exercises using Pure Geometry , but I will make an exception* . 😊 In right triangle ABD => tan⁡ϑ=BD/AB=2/4=1/2 => tanθ=1/2 (1) In right triangle ABC => tan⁡2ϑ=(x+2)/4⇒(2 tan⁡ϑ)/(1-tan²θ) = (x+2)/4 => (2⋅1/2)/(1-(1/2²) )=(x+2)/4 cause (1) => ……… x=10/3 etc

Let DE⊥AC (construction) and DC=x ,EC=y. Obviously orthogonal triangles ABD,ADE are equal. So AE=4 and DE=2. Area (ABC) =1/2 BC⋅AB=1/2 (x+2)⋅4=2x+4 (1) Also area (ABC) = area (ABD) + Area (ADE) + Area (DEC) =(1/2)⋅4⋅2+(1/2)⋅4⋅2+(1/2)⋅2y area (ABC) = 8+y (2) (1),(2)=> 2x+4=8+y => y=2x-4 (3) Pythagoras theorem in right triangle DEC => x²=y²+4 => x²=(2x-4)²+4 cause (3) At last x=10/3 or x=2 (it’s not valid. Why ???) substitute x=10/3 in (1) and Area (ABC)=2⋅10/3+4=32/3

Muito bom. Não é fácil resolver por 3 maneiras distintas.

Простенькая задача в 3- м методе доведена до абсурда , думал , что только у нас бывают - "из пустого в порожнее" . Ан нет , и там такие - порожняк любят гонять .

You dont have to draw any lines for another method. Once we know alpha, we can use the law of sines in triangle ADC.From ttriangle ABD we know the value of the sine and cosine theta. The rest is easy. This is also solvable using the angle bisector theorem.

Triangle ∆ABD: BD² + AB² = DA² 2² + 4² = DA² DA² = 4 + 16 = 20 DA = √20 = 2√5 As ∠ADC is an external angle to ∆ABD at D, ∠ADC = ∠DAB + ∠ABD = 90°+θ. Draw DE, where DE is perpendicular to BC and intersects CA. As ∠ADC = 90°+θ, ∠ADE = θ. Thus ∆DEA is an isosceles triangle as ∠ADE = ∠EAD = θ. Draw EF, where F is the midpoint of AD. As AD is the base of the isosceles triangle ∆DEA and E is the opposite vertex, EF bisects ∆DEA and creates two congruent right triangles ∆EFD and ∆AFE. As ∠EAF = ∠FDE = θ, both right triangles are similar to right triangle ∆ABD. Triangle ∆EFD: DE/FD = DA/AB DE/√5 = 2√5/4 = √5/2 DE = (√5)√5/2 = 5/2 As ∠EDC = ∠ABC = 90°, DC is colinear with BC, and ∠DCA is shared, ∆EDC and ∆ABC are similar. Let DC = x. Triangle ∆EDC: DC/ED = BC/AB x/(5/2) = (x+2)/4 4x = (5/2)(x+2) 4x = 5x/2 + 5 3x/2 = 5 x = (2/3)5 = 10/3 Triangle ∆ABC: Area = bh/2 A = (2+10/3)4/2 = (16/3)2 = 32/3 sq units

S=32/3≈10,67

I think that I have understood ALL three methods shown!

With bisector theorem 32/3

شكرا لكم على المجهودات يمكن استعمال AC^2=4^2+(2+DC)^2 2/DC =4/AC ..... DC=10/3 S(ABC)=32/3 الطريقة 2 tanBAD=1/2 tan(2BAD) =( 2 +DC)/4 = 4/3 ..... DC=10/3 S(ABC)=32/3

I got the third way. But the second method is the most fast and effective.