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Question Number 21571 by Tinkutara last updated on 27/Sep/17

A B C D is a cyclic quadrilateral; x, y, z

are the distances of A from the lines

B D, B C, C D respectively . Prove that

\frac{B D}{x} = \frac{B C}{y} + \frac{C D}{z} .

Answered by revenge last updated on 29/Sep/17

Commented by revenge last updated on 29/Sep/17

I t c a n b e p r o v e d t h a t :

(i) Δ A D E \sim Δ A B G

(i i) Δ A D F \sim Δ A C G

(i i i) Δ A F B \sim Δ A E C

F r o m (i), \frac{B G}{y} = \frac{D E}{z}

\Rightarrow \frac{B C}{y} + \frac{C D}{z} = \frac{G C - G B}{y} + \frac{C E + E D}{z} = \frac{G C}{y} + \frac{C E}{z} \dots (*)

F r o m (i i), \frac{C G}{y} = \frac{D F}{x} \dots (1)

F r o m (i i i), \frac{C E}{z} = \frac{F B}{x} \dots (2)

A d d i n g (1) & (2) w i t h (*), w e g e t

\frac{B C}{y} + \frac{C D}{z} = \frac{D F + F B}{x} = \frac{B D}{x}

Commented by Tinkutara last updated on 29/Sep/17

Thank you very much Sir!