in triangle ABC. AD drawn perpendicular to BC, BC:CD=3:1 .then prove that BC square (AB square -AC square)
Answers
Answer:
ANSWER
Equation of the circle is x
2
+y
2
=r
2
.
Let co-ordinates of the point P(x
1
,y
1
)=(rcosθ,rsinθ).
Since chord AB subtends right angle at the centre, therefore coordinates of point A(x
2
,y
2
)=(r,0) and point B(x
3
,y
3
)=(0,r).
We know that coordinates of the centroid of the △PAB
(h,k)=(
3
x
1
+x
2
+x
3
,
3
y
1
+y
2
+y
3
)=(
3
rcosθ+r+0
,
3
rsinθ+0+r
)
=(
3
1
(r+rcosθ),
3
1
(r+rsinθ))
Therefore, h=
3
1
(r+rcosθ)⇒h−
3
1
r=
3
1
rcosθ ...(i)
and k=
3
1
(r+rsinθ)⇒k−
3
1
r=
3
1
rsinθ ...(ii)
Squaring and adding Eqs. (i) and (ii), we get
(h−
3
1
r)
2
+(k−
3
1
r)
2
=
9
1
r
2
(cos
2
θ+sin
2
θ)=
9
1
r
2
Thus, the locus of the centroid of △PAB is (x−
3
r
)
2
+(y−
3
r
)
2
=
9
r
2
, which is a circle.
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