Q17
Q17: "In △ABC, ∠A = 70°, D is on the side AC, and the angle bisector of ∠A intersects BD at H such that AH : HE = 3 : 1 and BH : HD = 5 : 3. Then ∠C in degrees is" Answer: 55 Solution: We use Angle Bisector Theorem and Mass points here. mA = 1 and mE = 3 => mH = 4. mB/mD = HD/BD = 3/5. Using mH = 4 we get mB = 3/2 and mD = 5/2. mC = mD - mA = 3/2. Using angle bisector theorem: AB/AC = EB/EC = mC/mB = (3/2)/(3/2) = 1 => AB = AC So ABC is iscosceles. Angle C = Angle B = (180-70)/2 = 55.