Can You Find θ? Most Get This Geometry Problem Wrong | Geometry

[00:00:00]Hi, welcome to my channel. In this video, we want  to solve the following problem. In triangle ABC, ∠ABC = 15° and ∠BCA = 30°. Let D be on BC such  that BD = DC. If ∠ADC = θ, the goal is to find θ.

[00:00:27]And now for the solution. Let BD = DC = a. Next,  from A, draw a line meeting BD at P such that ∠APD = 30°. Now consider triangle ABP and segment  PD. ∠APD is an exterior angle of triangle ABP.

[00:00:56]Since an exterior angle equals the sum of interior  opposite angles, ∠PAB = 30° - 15° = 15°. Since ∠ABP = ∠PAB = 15°, triangle ABP is isosceles, so  AP = BP. Let AP = BP = b, then PD = a - b. Next, consider triangle APC and segment AD. Since  ∠APC = ∠PCA = 30°, triangle APC is isosceles, so AC = AP = b. Next, drop AQ perpendicular  to PC with Q on PC. In an isosceles triangle, the height from the vertex bisects the base, so PQ  = QC. Therefore, QC = ((a - b) + a)/2 = a - b/2.

[00:02:10]Then DQ = a - (a - b/2) = b/2. Next consider  triangle AQC. Since the triangle is right-angled, sin30° = AQ/AC. Using sin30° = 1/2,  we have AQ/b = 1/2 so AQ = b/2.

[00:02:38]Finally consider triangle ADQ. Since DQ  = AQ = b/2, triangle ADQ is isosceles, so the base angles are equal. Thus θ = (180°  - 90°)/2. Therefore, θ = 45° which is the answer. This marks the end of the video. If  you enjoyed this video, please leave a like and subscribe to my channel for more content.  Thanks for watching and see you all next time.