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 = ...

 Rectangle properties: Problems: Prove that medians/altitudes/angle bisectors of a triangle are concurrent. For altitudes: Solution: If you apply the formula for CosA and CosC this problem is effectively to prove that c.CosA = b.CosC. but c.CosA = AH So pr. that AH = b.CosC but CosC = CH/a So pr. that. AH/CH = b/a By Angle Bisector theorem b/a = AD/BD. So pr. that AH/CH = AD/BD. By Ceva's theorem AD/BD * BM/CM * CH/AH = 1 But BM = CM So AD/BD = AH/CH Hence proved.

  Question 1 Solution: ACQD is ||gram coz AD || CQ and AD = CQ. So [DPQ] = [APC] from congruence. [APC] = [BPC] coz same base(PC) and height from opposite side. So [DPQ] = [BPC]. Question 2 Solution: [LZY] = [XYZ] + [TXZ] + [LTX] [MZYX] = [XYZ] + [TXZ] + [MTZ] So we need to prove that [LTX] = [MTZ]. LMZX is a trapezium so triangles between parallel sides are similar and other 2 have equal area which can be proven using side ratios. Which gives [LTX] = [MTZ]. I am not sure what was the use of LX = XY = YN here? Question 3 Solution: [ADF] = [BDF] = 3 coz same base (DF) and height. [FCE] = [ADF]  = 3due to AAS congruence. (AD = CE, F is vertical angle, angle DAF = angle CEF). [FCB] = [FCE] = 3 coz same base(BC = EC) and height (from F to BE). Diagonal BD splits ||gram ABCD in 2 equal areas, prove by congruence. So [BDF] + [BFC] = 1/2[ABCD] = 3 + 3 So [ABCD] = 12. Question 4 Solution : Can be easily proven that [RDF] = [RAS] = [BSP] = [PFC] = k/8 where k  = [ABCD]. Hint: ASR ...