In the following figure, ABD is a triangle right angled at A and AC �

1.	In the following figure, ABD is a triangle right angled at A and AC ⊥BD. Show that (i) AB2= BC.BD (ii) AC2 = BC.DC (iii) AD2 = BD.CD
Answer» Data: In ∆ABD, ∠A = 90°, AC ⊥ BD. To Proved: AB² = BC.BD (ii) AC² = BC.DC iii) AD² = BD.CD (i) AB² = BC.BD ∆ACB ~ ∆BAD (. Theorem7) ∴ \(\frac{AB}{BD} = \frac{BC}{AB}\) ∴ AB²= BC × BD. (ii) AC² = BC.DC ∆BCA ~ ∆ACD ∴ \(\frac{AB}{AD} = \frac{AC}{CD} = \frac{BC}{AC}\) ∴ AC × AC = BC × CD ∴ AC² = BC × CD (iii) AD² = BD.CD ∆ACD ~ ∆BAD ∴ \(\frac{AD}{BD} = \frac{CD}{AD} = \frac{AC}{AB}\) ∴ AD × AD = BD × DC ∴ AD² = BD × DC

In the following figure, ABD is a triangle right angled at A and AC ⊥BD. Show that (i) AB2= BC.BD (ii) AC2 = BC.DC (iii) AD2 = BD.CD

Answer»

Data: In ∆ABD, ∠A = 90°,

AC ⊥ BD.

To Proved: AB² = BC.BD

(ii) AC² = BC.DC

iii) AD² = BD.CD

(i) AB² = BC.BD

∆ACB ~ ∆BAD (. Theorem7)

∴ \(\frac{AB}{BD} = \frac{BC}{AB}\)

∴ AB²= BC × BD.

(ii) AC² = BC.DC

∆BCA ~ ∆ACD

∴ \(\frac{AB}{AD} = \frac{AC}{CD} = \frac{BC}{AC}\)

∴ AC × AC = BC × CD

∴ AC² = BC × CD

(iii) AD² = BD.CD

∆ACD ~ ∆BAD

∴ \(\frac{AD}{BD} = \frac{CD}{AD} = \frac{AC}{AB}\)

∴ AD × AD = BD × DC

∴ AD² = BD × DC