In triangle ABC , AB=AC and D is a point on AC such that BC×BC=AC×CD prove that BD=BC
rehaanboby2003:
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Given
In
Δ
A
B
C
A
B
=
A
C
and
D
is a point on
A
C
such that
B
C
×
B
C
=
A
C
×
A
D
We are to prove
B
D
=
B
C
Proof
Rearrenging the given relation
B
C
×
B
C
=
A
C
×
A
D
We can write
B
C
C
D
=
A
C
B
C
→
Δ
A
B
C
similar
Δ
B
D
C
Their corresponding angle pairs are:
1
.
∠
B
A
C
= corresponding
∠
D
B
C
2
.
∠
A
B
C
= corresponding
∠
B
D
C
3
.
∠
A
C
B
=corresponding
∠
D
C
B
So as per above relation 2 we have
∠
A
B
C
=
corresponding
∠
B
D
C
Again in
Δ
A
B
C
A
B
=
A
C
→
∠
A
B
C
=
∠
A
C
B
=
∠
D
C
B
∴
In
Δ
B
D
C
,
∠
B
D
C
=
∠
B
C
D
→
B
D
=
B
C
Alternative way
The ratio of corresponding sides may be written in extended way as follows
B
C
C
D
=
A
C
B
C
=
A
B
B
D
From this relation we have
A
C
B
C
=
A
B
B
D
⇒
A
C
B
C
=
A
C
B
D
→
As
A
B
=
A
C
given
⇒
1
B
C
=
1
B
D
⇒
B
C
=
B
D
Proved
Hope, this will help
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