4) In A ABC, ZABC = 90, point P and Q are midpoints of AB
and BC respectively. Show that AQ2 + CP2 = 4xPQ2
Answers
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
Answer
AQ
2
=AB
2
+BQ
2
=AB
2
+(
2
BC
)
2
=AB
2
+
4
BC
2
...(i)
Similarly CP
2
=BC
2
+BP
2
=BC
2
+(
2
AB
)
2
=
4
AB
2
+BC
2
...(ii)
∴AQ
2
+CP
2
=AB
2
(1+
4
1
)+BC
2
(1+
4
1
)
=
4
5
(AB
2
+BC
2
)=
4
5
AC
2