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