(Q14) ABD is a triangle right angled at A and AC ⊥ BD.Show that

(i) AB² = BC.BD

(ii) AD² = BD.CD

(iii) AC² = BC.DC

Solution :

Given :In ∆ABD; ∠A = 90° AC ⊥ BD

R.T.P :(i) AB² = BC.BD

Proof :In ∆ABD and ∆CAB,

∠BAD = ∠ [each 90°]

∠B = ∠B [common]

∴ ∆ABD ∼ ∆

[by A.A. similarity condition]

[∵ Ratios of corresponding sides of similar triangles are equal]

⇒ AB. = BC.

AB² = .

ii) AD² = BD.CD

Proof : In ∆ABD and ∆CAD

∠BAD = ∠ [each 90°]

∠D = ∠ (common)

∴ ∆ABD ∼ ∆ [A.A similarity]

AD.AD = .

= BD.CD

iii) AC² = BC.DC

Proof : From (i) and (ii)

∆ACB ∼ ∆

[∵ ∆BAD ∼ ∆BCA ∼ ∆ACD]

AC.AC = .

= BC.DC