To Prove : △ADE∼△ACB
Proof :
(i) In △ADE and △ACB
(1) ∠A=∠A [common]
(2) ∠AED=∠ABC=90o (given)
∴ △ADE∼△ACB [AA axiom]
(ii) (AC)2=(AB)2+(BC)2
169=(AB)2+25
AB=12cm
∵ △ADE∼△ACB
∴ DEBC=ADAC=AEAB
∴ DEBC=AEAB
DE5=412
DE=2012=53=123cm
Now, ADAC=AEAB
AD13=412
AD=13×412=133=413cm.
(iii) Ar.of(△ABC)Ar.of(△ADE)=AB2AE2=14416=91
Ar.of(△ADE)+Ar.of(BCED)Ar.of(△ADE)=9
1+Ar.of(BCED)Ar.of(△ADE)=9
Ar.of(△ADE)Ar.of(BCED)=18