Questions 26 please help me
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ΔABD similar to ΔPQD
∴x/z=BD/QD
1/x =QD/z.BD---------1
ΔCDB similar to ΔPQB
∴y/z=BD/BQ
1/y =BQ/z.BD-----------2
adding 1 and 2
1/x+1/y=(BQ+QD)/z.BD
⇒BD/z.BD⇒1/z
1/x +1/y =1/z
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