Question

∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC × BD

(ii) AC² = BC × DC

(iii) AD² = BD × CD

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

Answer 1

(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"`

⇒ AB² = 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"`

⇒ AC² = DC × BC

(iii) In ΔDCA and 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"`

⇒AD² = 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