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

Given, ABCD is a right angled triangle and AC is perpendicular to BD

(i) consider two triangles ACB and DAB

⇒ We have ∠ ABC = ∠ DBC

⇒ ∠ ACB = ∠ DAB

⇒ ∠ CAB = ∠ ADB

∴ they are similar and corresponding sides must be proportional

i.e, ∠ ADC = ∠ ADB

⇒

∴ AB² = BC × CD

(ii) ∠ BDA = ∠ BDC = 90°

⇒ ∠ 3 = ∠ 2 = 90° ∠ 1

⇒ ∠ 2 + ∠ 4 = 90° ∠ 2

⇒ From AAA criterion of similarity we have in two triangles if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similar

Their corresponding sides must be proportional

⇒

⇒

⇒ = BC × DC

(iii) In two triangles ADB and ABC we have

∠ ADC = ∠ ADB

⇒ ∠ DCA = ∠ DAB

⇒ ∠ DAC = ∠ DBA

⇒ ∠ DCA = ∠ DAB

⇒ Triangle ADB and ABC are similar and so their corresponding sides must be proportion.

⇒ = =

⇒ =

⇒ AD² = DB × DC

Hence proved