The-triangle-ABC-has-CA-CB-P-is-a-point-on-the-circumcircle-between-A-and-B-and-on-the-opposite-side-of-the-line-AB-to-C-D-is-the-foot-of-the-perpendicular-from-C-to-PB-Show-that-PA-PB-2-PD

Question Number 17614 by Tinkutara last updated on 08/Jul/17

The triangle ABC has CA = CB . P is a

point on the circumcircle between A

and B (and on the opposite side of the

line AB to C) . D is the foot of the

perpendicular from C to PB . Show that

PA + PB = 2 \cdot PD .

Commented by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

Commented by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

C A \times P B + C B \times P A = C P \times A B ({p e t o l e m y}^{'} s t e o r e m)

\Rightarrow C A (P A + P B) = C P \times A B

\Rightarrow P A + P B = \frac{C P \times A B}{C A}

\frac{A B}{A C} = \frac{s i n C}{s i n B} = \frac{s i n (180 - 2 B)}{s i n B} = \frac{s i n 2 B}{s i n B} = 2 c o s B

c o s ∡ C P B = \frac{P D}{C P} \Rightarrow C P = \frac{P D}{c o s ∡ C P B} = \frac{P D}{c o s A} =

= \frac{P D}{c o s B}

\Rightarrow P A + P B = \frac{C P . A B}{C A} = 2 c o s B . \frac{P D}{c o s B} = 2 . P D .

n o t e : ∡ C P B = ∡ C A B, ∡ A P C = ∡ A B C

Commented by Tinkutara last updated on 09/Jul/17

Thanks Sir!