Math Olympiad | A Nice Geometry Problem | 2 Different Methods 

From one of the comments, I've learned a method which is really thinking outside the box. I would like to present it in a more formal way. Method using constructed congruent triangles, circle properties: 1. In triangle ABC, angle C = 180 - 20 - 80 = 80 = angle A. Triangle ABC is isosceles trianlge (base angles equal). Hence AB = BC. 2. Use B as centre to draw a circle with radius= BC, hence passing through B and A (as AB=BC) and a point E on circumference beyond arc CA such that line CE = radius of circle. An equilateral triangle BCE with sides as radii is formed with angles = 60. 3. Triangle ACE is congruent to triangle DBA (ASA): Angle ACE = 80 - 60 = 20 (subtracting angle of equilateral triangle) Angle CEA = 10 (angle at circumference is half of angle at centre which is 20) CE = AB (radii of circle by construction) Hence x = AC = DB = 6 (corresponding sides of congruent triangles)

1st solution is really Mind blowing.

We can also solve this way: Let a point E on AB such that DE=AE so in this way angle DEB =20°; That means BD=DE=AE=6; in isosceles ∆AED,AD=12Cos10°; Drop a perpendicular AF(let F be a point on CD) on CD then in ∆AFD,then AF=ADSin30°=6Cos10°; now in ∆ACF(Angle CAF=10°) xCos10°=AF=6Cos10°; So x=6.

Subscribed and grateful!

Good content, I often train myself with these videos you uploaded 👍👍

To a point E above AB, draw an Equilateral triangle EBC. A circle passes through C, A, and E with B as its center. Hence ∠AEC = 10° and ∠ACE =20°(angles subtended on the circumference) CE= AB= Radius. Hence ∆BAD is Congruent to ∆AEC by ASA. AC = BD = 6

Summary of the methods explained: exterior angle theorem is used to find angles in the second triangle. Then using the definition of vertical angles, a right triangle is formed. Then the isosceles triangle is formed justifying the angle bisector theorem and two angles equaling 80 degrees are justified so that two sides are set equal. The second method involves Law of Sines and sets two involved sides equal. I could be wrong.

Used the second method

6/ sin 10 =y / sin 20 X/ sin 30 = y /sin 80 2 x = y / cos 10 ........=6 sin 20/ sin10 × 1/ cos10 =12 sin 20 / sin 20 .........= 12 X= 6...?

Difícil. 1o método mais elegante.

First like,first comment.thanks sir