In a parallelogram ABCD, a point P is taken on AD such that AP : AD is 1:n. if BP intersects AC at Q, then AQ : AC is
A) 2/(n+1)
B) 1/(n-1)
C) 1/(n+1)
D) 2/(n-1)
Answer
659.4k+ views
Hint: We know that in a parallelogram opposite sides are parallel and equal in length. We will use this property of parallelogram and angle-side-angle congruent theorem of triangle to solve this problem.
Complete step-by-step answer:
Representing the data given in the above question into a diagram
Here ABCD is a parallelogram and AC is one of its diagonal.
P is a point on AD such that AP : AD is 1:n and BP intersects AC at Q
Let B’ be a point on BC and B’D intersects AC at Q’ and B’D is parallel to the BP
In the figure, \[PQ||Q'D\] and \[BP||B'D\]
So in ADQ’ triangle, $\dfrac{{AQ}}{{AQ'}} = \dfrac{1}{n}$ ………………(1)
Similarly in BCQ triangle, $\dfrac{{CQ'}}{{CQ}} = \dfrac{{CB'}}{{CB}}$
As ABCD and BPDB’ are parallelogram,
So, AD=BC
PD=BB’
And AP=CB’
As triangle APQ and triangle CB’Q’ are congruent as per angle-side-angle theorem,
Hence AQ=CQ’………….(2)
From equation (1) we have,
$\dfrac{{AQ}}{{AQ'}} = \dfrac{1}{n}$
By dividing one fraction by another we get,
$\dfrac{{AQ'}}{{AQ}} = n$
Adding 1 on both side of the above equation we get,
$\dfrac{{AQ'}}{{AQ}} + 1 = n + 1$
Simplifying the above equation we get,
$\dfrac{{AQ' + AQ}}{{AQ}} = n + 1$
Putting equation 2 in the above equation we get,
$\dfrac{{AQ' + CQ'}}{{AQ}} = n + 1$
As we know that, AQ’+CQ’ is AC, then substituting it in the above equation we get,
$\dfrac{{AC}}{{AQ}} = n + 1$
Flipping the terms of both of the side upside down,
$\dfrac{{AQ}}{{AC}} = \dfrac{1}{{n + 1}}$
Therefore AQ : AC is 1/(n+1)
Hence the option C is correct.
Note: In a parallelogram, opposite sides are parallel and equal with each other. Similarly opposite angles are also equal.
In the angle-side-angle congruent theorem of triangle, it is stated that if two angles and the included side of a triangle is equal with the same of another triangle, then the triangles are congruent.
You need to remember the theorem and properties of geometric shapes.
Complete step-by-step answer:
Representing the data given in the above question into a diagram
Here ABCD is a parallelogram and AC is one of its diagonal.
P is a point on AD such that AP : AD is 1:n and BP intersects AC at Q
Let B’ be a point on BC and B’D intersects AC at Q’ and B’D is parallel to the BP
In the figure, \[PQ||Q'D\] and \[BP||B'D\]
So in ADQ’ triangle, $\dfrac{{AQ}}{{AQ'}} = \dfrac{1}{n}$ ………………(1)
Similarly in BCQ triangle, $\dfrac{{CQ'}}{{CQ}} = \dfrac{{CB'}}{{CB}}$
As ABCD and BPDB’ are parallelogram,
So, AD=BC
PD=BB’
And AP=CB’
As triangle APQ and triangle CB’Q’ are congruent as per angle-side-angle theorem,
Hence AQ=CQ’………….(2)
From equation (1) we have,
$\dfrac{{AQ}}{{AQ'}} = \dfrac{1}{n}$
By dividing one fraction by another we get,
$\dfrac{{AQ'}}{{AQ}} = n$
Adding 1 on both side of the above equation we get,
$\dfrac{{AQ'}}{{AQ}} + 1 = n + 1$
Simplifying the above equation we get,
$\dfrac{{AQ' + AQ}}{{AQ}} = n + 1$
Putting equation 2 in the above equation we get,
$\dfrac{{AQ' + CQ'}}{{AQ}} = n + 1$
As we know that, AQ’+CQ’ is AC, then substituting it in the above equation we get,
$\dfrac{{AC}}{{AQ}} = n + 1$
Flipping the terms of both of the side upside down,
$\dfrac{{AQ}}{{AC}} = \dfrac{1}{{n + 1}}$
Therefore AQ : AC is 1/(n+1)
Hence the option C is correct.
Note: In a parallelogram, opposite sides are parallel and equal with each other. Similarly opposite angles are also equal.
In the angle-side-angle congruent theorem of triangle, it is stated that if two angles and the included side of a triangle is equal with the same of another triangle, then the triangles are congruent.
You need to remember the theorem and properties of geometric shapes.
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