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.

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