#### Question

∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that

(i) AB^{2} = BC x BD

(ii) AC^{2} = BC x DC

(iii) AD^{2} = BD x CD

(iv) `"AB"^2/"AC"^2="BD"/"DC"`

#### Solution

(i) In ΔADB and ΔCAB

∠DAB = ∠ACB = 90°

∠ABD = ∠CBA (common angle)

∠ADB = ∠CAB (remaining angle)

So, ΔADB ~ ΔCAB (by AAA similarity)

Therefore `"AB"/"CB"="BD"/"AB"`

⇒ AB^{2} = CB × BD

(ii) Let ∠CAB = x

In ΔCBA

∠CBA = 180° − 90° − 𝑥

∠CBA = 90° − 𝑥

Similarly in ΔCAD

∠CAD = 90° − ∠CAD = 90° − 𝑥

∠CDA = 90° − ∠CAB

= 90° − 𝑥

∠CDA = 180° −90° − (90° − 𝑥)

∠CDA = x

Now in ΔCBA and ΔCAD we may observe that

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA = 90°

Therefore ΔCBA ~ ΔCAD (by AAA rule)

Therefore `"AC"/"DC"="BC"/"AC"`

⇒ AC^{2} = DC × BC

(iii) In ΔDCA & ΔDAB

∠DCA = ∠DAB (both are equal to 90°)

∠CDA = ∠ADB (common angle)

∠DAC = ∠DBA (remaining angle)

ΔDCA ~ ΔDAB (AAA property)

Therefore `"DC"/"DA"="DA"/"DB"`

⇒AD^{2} = BD × CD

(iv) From part (i) 𝐴𝐵2 = 𝐶𝐵 × 𝐵𝐷

From part (ii) 𝐴𝐶2 = 𝐷𝐶 × 𝐵𝐶

Hence `"AB"^2/"AC"^2=(CBxxBD)/(DCxxBC)`

`"AB"^2/"AC"^2="BD"/"DC"`

Hence proved