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Triangle \( A B C \) has a right angle at \( C \). The bisector of \( \angle B C A \) meets \( A B \) at \( D \), and \( D E \) is the perpendicular from \( D \) to AC. Prove that \( 1 / B C+1 / A C=1 / D E \).

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DE \(\perp\) AC

In right angles ΔAED & Δ ACB,

∠AED = ∠ACB = 90°

∠DAE = ∠BAC (common angles)

\(\therefore\) Δ AED ~ Δ ACB (By AA similarity)

\(\therefore\) \(\frac{AE}{AC} = \frac{AD}{AB} = \frac{DE}{BC}\)---(1)

By angle bisector theorem, we have

\(\frac{AC}{BC}=\frac{AD}{BD}\)

⇒ AC = \(\frac{BC.AD}{BD}\)---(2)

Now, 

\(\frac{DE}{BC}+\frac{DE}{AC} = \frac{DE}{BC} + \frac{DE.BD}{BC.AD}\) (From 2)

\(\frac{DE}{BC}(1+\frac{BD}{AD})\) 

=  \(\frac{DE}{BC}(\frac{AD+BD}{AD})\)

\(\frac{DE}{BC}\times\frac{AB}{AD}\)

\(\frac{DE}{BC}\times\frac{BC}{DE}\) = 1 (From 1) 

⇒ \(\frac{DE}{BC}\times\frac{BC}{DE}\) = 1

⇒ \(\frac1{BC}+\frac1{AC} = \frac1{DE}\)

Hence proved

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