Problem:
Triangle ABC has AC=3,BC=4, and AB=5. Point D is on AB, and CD bisects the right angle. The inscribed circles of β³ADC and β³BCD have radii raβ and rbβ, respectively. What is raβ/rbβ?
Answer Choices:
A. 281β(10β2β)
B. 563β(10β2β)
C. 141β(10β2β)
D. 565β(10β2β)
E. 283β(10β2β) Solution:
By the Angle Bisector Theorem,
AD=5β 3+43β=715β and BD=5β 3+44β=720β
To determine CD, start with the relation Area (β³ADC)+Area(β³BCD)= Area (β³ABC) to get
22β3β CDβ+22β4β CDβ=23β 4β
This gives CD=7122ββ. Now use the fact that the area of a triangle is given by rs, where r is the radius of the inscribed circle and s is half the perimeter of the triangle. The ratio of the area of β³ADC to the area of β³BCD is the ratio of the altitudes to their common base CD, which is BDADβ=43β. Hence