Question

3 IF A-B-C and LAB= 16, UCBC-5, then find 21C).​

Answer 1

ACD BPD, CDP ADB which means I. PD/PB = CD/CA

III PD/PC = BD/AB.

1 PD/PC = 1 - BD/AB(by II)

= (AB-BD)/AB

= (BC-BD)/AB (since triangle AABC is equilateral) = CD/CA (because triangle ABC is equilateral) = PD/PB (from I)

∴ 1 - PD/PC = PD/PB.

PD to both sides On dividing we get

1/PD- 1/PC = 1/PB.

Moving 1/PC to the right gives us the desired equality.