AABC is a right angled triangle. 2B = 90° a perpendicular line from B to AC intersect AC at D. Prove that AABC, AADB and ABCD are similar to each other.
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Answer:
To Prove : △ADE∼△ACB
Proof :
(i) In △ADE and △ACB
(1) ∠A=∠A [common]
(2) ∠AED=∠ABC=90
o
(given)
∴ △ADE∼△ACB [AA axiom]
(ii) (AC)
2
=(AB)
2
+(BC)
2
169=(AB)
2
+25
AB=12cm
∵ △ADE∼△ACB
∴
BC
DE
=
AC
AD
=
AB
AE
∴
BC
DE
=
AB
AE
5
DE
=
12
4
DE=
12
20
=
3
5
=1
3
2
cm
Now,
AC
AD
=
AB
AE
13
AD
=
12
4
AD=
12
13×4
=
3
13
=4
3
1
cm.
(iii)
Ar.of(△ADE)
Ar.of(△ABC)
=
AE
2
AB
2
=
16
144
=
1
9
Ar.of(△ADE)
Ar.of(△ADE)+Ar.of(BCED)
=9
1+
Ar.of(△ADE)
Ar.of(BCED)
=9
Ar.of(BCED)
Ar.of(△ADE)
=
8
1
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