At 8:35, Note that ΔADE is 45°-45°-90°, therefore an isosceles right triangle, and DE = AE, Also note that ΔABE and ΔABC are similar by angle-angle (common angle Θ and their right angles). By ratios of sides of similar triangles, AE is twice as long as BE. So, if BE is assigned a length y, AE and DE have length 2y. X = BD = BE + DE = y + 2y = 3y. ΔACE is also similar to ΔABC by angle-angle (common angle

Once drawn height AE, being AED right with angle in D = 45° this means it is isosceles. Then AE = ED = y Triangles AEC and ABC are similar being both right and having angle in C in common, so we can write: a : y = 2a : (y+2) 2ay = ay + 2a a is a common factor that can be semplified y = 2 = AH = HD then applying Euclid's theorem: AH² = BE*EC 2² = BE*(2+2) BE = 4/4 = 1 x = BE + ED = 1 + 2 = 3

A shortcut in first method: Notice that So quadrilateral ABDE is inscribed in a circle => CE•AC = CD • BC => a•2a = 2•(x+2) => a² = x +2 (1) ………….. Then Math Booster goes on …… Another solution Obviously a²=y²+2² => x +2 = y²+4 , cause (1) x= y²+2 (2) Area ABC = Area ABE + Area EBC => a•2a =a ² + y(x+2) => a²=(x+2)y x+2= (x+2)y => (x+2)(1-y)=0 => y=1 (2)=> x=1²+2 => x=3

Another solution: tang(ACB)=a/2a=1/2 in triangle ABC, AH is perpendicular from A to BC =>in triangle AHD angle(AHD)=90 =>triangle AHD is isosceles (AH=HD)=> tang(ACH)=AH/HC=AH/(HD+2)=1/2=>AH=HD=2 tang(ACB)=sin(ACB)/cos(ACB)=sin(ACB)/SQRT(1-(sin^2(ACB))=1/2 => sin(ACB)=SQRT(1/5) angle(ACB)=angle(BAH)=>sin(ACB)=sin(BAH)=BH/AB=(BD-HD)/AB=(x-2)/a ===>>>(x-2)/a=SQRT(1/5)===>>>a=(x-2)*SQRT(1/5) in triangle ABC Pitagora theorem (x+2)^2=a^2+(2*a)^2=5*a^2 in this equation insert a=(x-2)*SQRT(1/5) and the result is a second grade equation : ====> 3*x^2-13*x+12=0

Arctangent 2 is 63.43°, arctangent 1/2 26.57°. Angle C-A-D is 18.43°. 2/sin18.43°=2a/sin135°=4.47 and a=2.24, by Pythagorean theorem x+2=5 therefore x=3.

Very good!!

Here's another solution. (Sorry about my poor English) Let's set E on AC, satisfying DE⊥BC. Since angle BAE+EDB=180°, quadrangle ABDE is inscribed in a circle. Now we can find that angle ADB=ADE=45°. Since two angles of circumference are same, AB=AE. Therefore, E is the midpoint of AC. Let's set H on BD, satisfying AH⊥BD. We can easily find that triangle AHC is similar to EDC. DH=AH=2. Since triangle ABC is similar to HBA, BH=1. Therefore, BD=BH+DH=1+2=3

tan(alpha) = 2 From law of sines and trigonometric identities express lengths in obtuse triangle in terms of A From law of cosine with angle 45 we get one equation in terms of a and x From Pythagorean theorem (which is special case of cosine law) we get second equation in terms of a and x and we have quadratic equation Finally i have got two solutions x = 1/2 and x=3 and i dont know which one is extraneous

Intanto 5a^2=(x+2)^2...x=√5a-2.uso sempre il teorema dei seni,mi appoggio ad α=BAD..risulta α=arctg3..a=sin135/cosα=√5...quindi,in sintesi,x=√5√5-2=3