Problem:
In rectangle ABCD,AD=1,P is on AB, and DB and DP trisect ∠ADC. What is the perimeter of △BDP?
Answer Choices:
A. 3+33
B. 2+343
C. 2+22
D. 23+35
E. 32+53
Solution:
Both triangles APD and CBD are 30−60−90∘ triangles. Thus DP=323 and DB=2. Since ∠BDP=∠PDB, it follows that PB=PD=323. Hence the perimeter of △BDP is 323+323+2=2+343.
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