In Fig. 6.53, ABD is a triangle right angled at Aand AC ⊥ BD. Show that.

: (i) In ΔADB and ΔCAB, we have

∠DAB = ∠ACB    (Each of 90^o)

∠ABD = ∠CBA     (Common angle)

Therefore,

ΔADB~ ΔCAB   (AA similarity)

AB² = CB * BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180^o – 90^o – x

∠CBA = 90^o – x

Similarly, in ΔCAD

∠CAD = 90^o - ∠CBA

= 90^o – x

∠CDA = 180^o – 90^o – (90^o – x)

∠CDA = x

In triangle CBA and CAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA     (Each 90^o)

Therefore,

ΔCBA ~ΔCAD     (By AAA similarity)

AC² = DC * BC

(iii) In ΔDCA and ΔDAB, we have

∠DCA = ∠DAB     (Each 90^o)

∠CDA = ∠ADB     (Common angle)

Therefore,

ΔDCA~ ΔDAB     (By AA similarity)

AD² = BD * CD