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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