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Standard IX
Mathematics
Theorem 1: Parallelograms
Question 9 ii...
Question
Question 9 (ii)
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
AP = CQ
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Solution
In
Δ
A
P
D
a
n
d
Δ
C
Q
B
,
∠
A
D
P
=
∠
C
B
Q
(Alternate interior angles)
AD =CB (Opposite sides of parallelogram ABCD)
DP=BQ (Given)
∴
Δ
A
P
D
≅
Δ
C
Q
B
(using SAS congruence rule)
As we have proved that triangle APD is congruent to triangle CQB,
Hence, AP= CQ (CPCT)
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Similar questions
Q.
Question 9 (ii)
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
AP = CQ
Q.
Question 9 (iv)
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
AQ = CP
Q.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
AQ = CP
Q.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
AP = CQ
Q.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ (see the given figure) Show that:
Δ
A
Q
B
≅
Δ
C
P
D
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Theorem 1: Parallelograms
Standard IX Mathematics
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