Math IOQM

Q29 - In quadrilateral ABCD, ∠DAB = ∠ABC = 110º, ∠BCD = 35º, ∠CDA = 105º and AC̅ bisects ∠DAB. Find ∠ABD. Answer: 40 degree. Solution:   Quick Solution: Let I denote the incenter of △ABD. Then quadrilateral IBCD is cyclic since ∠DIB = 90º + ½ ∠DAB = 145º. Hence we obtain ∠IBD = ∠ICD = 180º − (55º + 105º) = 20º and ∠ABD = 40º. Detailed solution: Step 1: If I is the incenter then ∠DIB = 90º + ½ ∠DAB = 145º. Why? Take any triangle XYZ. Let's say the angle bisectors of X and Z meet at O.  Then prove that ∠XOZ = 90º + ½ ∠XYZ. It's quite simple. Since I is the incenter here it means all the angle bisectors of A,B,D in △ABD meet at I. So this result will be easy to prove. Step 2: IBCD is a cyclic quadrilateral. Why? We just need to prove that 1 pair of opposite angles sums up to 180º. Since ∠I = 145º and ∠I = 35º, it's proved. Step 3: ∠IBD = ∠ICD why? Since in a circle angles subtended by the same chord are equal and both these angles are subtended by the chord ID. Now ∠ICD = ...

14. \( O \) and \( I \) are the circumcentre and incentre of \( \triangle ABC \) respectively. Suppose \( O \) lies in the interior of \( \triangle ABC \) and \( I \) lies on the circle passing through \( B, O, \) and \( C \). What is the magnitude of \( \angle BAC \) in degrees? You can apply a trick here. If you can construct a triangle which satisfies the conditions given in the question, you can answer using that. Here if you construct an equilateral triangle, O will lie in the interior of the triangle and I will lie on the circle passing through O,B,C since in the equilateral triangle O and I are same. So the answer will simply be 60 degrees. Detailed solution copied from here : We know, because \( I \) is the incenter, \[ \angle IBD = \frac{B}{2}, \quad \angle ICD = \frac{C}{2} \] \[ \Rightarrow \angle BIC = \pi - \left( \frac{B}{2} + \frac{C}{2} \right) = \pi - \left( \frac{\pi - A}{2} \right) = \frac{\pi}{2} + \frac{A}{2} \] But, \( \angle BIC = \angle BOC \), because c...