Quadrilaterals: class 9 math: 9 Math

NCERT Exercise 8.2:Part-2: 9th math

Quadrilaterals Class 9 math NCERT Exercise 8.2 Question (5) In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Solution

Given, ABCD is a parallelogram.

E and F are mid-points of sides AB and CD respectively.

Therefore, to prove that

Line segments AF and EC trisect the diagonal BD.

That is to prove that DP = PQ = QB

Since, F is the mid-point of DC

Therefore, DF = FC

Similarly, E is the mid-point of AB

Therefore, AE = EB

Now, since ABCD is a parallelogram

Therefore, DC||AB

Therefore, FC ||AE

Now, since, FC||AE and FC = AE

Therefore, quadrilateral AECF is a parallelogram.

[Because if one pair of opposite sides of a quadrilateral are parallel and equal, then it is a parallelogram]

Therefore, AF||EC

[Since, AEFC is a parallelogram]

Now, in triangle ABP,

E is the mid-point of side AB [As given in question]

And, EQ||AP

[Because AF||EC]

Therefore, according to Mid-Point Theorem which says that if a line is drawn through the mid-point of any side of a triangle and parallel to another side, it bisects the third side.

Therefore, Q is the mid-point of BP

This means, QB = PQ - - - - - (i)

Now, in triangle QCD,

F is the mid-point of side DC [As given in question]

And, PF||QC

[Because AF||EC (opposite sides of parallelogram)]

Therefore, according to Mid-Point Theorem which says that if a line is drawn through the mid-point of any side of a triangle and parallel to another side, it bisects the third side.

Therefore, P is the mid-point of DQ

This means, DP = PQ - - - - - (ii)

Now, from equation (i) and equation (ii), we get

DP = PQ = QB

Thus, segment AF and EC trisect the diagonal BD Proved

Quadrilaterals Class 9 math NCERT Exercise 8.2 Question (6) Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution

Let, ABCD is the given quadrilateral

And, E, F, G and H are mid-points of side AB, BC, CD and AD

Thus, to prove that HF and EG bisect each other.

Let join points A and C

In triangle ACD,

H and G are mid-points of side AD and DC

[As given]

Now, according to the Mid-point Theorem which says that the line segment joining the mid-points of any two sides of a triangle is parallel and half to the third side.

Therefore, HG||AC - - - - - (i)

And, HG = `1/2`AC - - - - - (ii)

Now, in triangle ABC

E and F are mid-points of side AB and BC [As given]

Therefore according to Mid-Points Theorem

EF || AC - - - - - (iii)

And, EF = `1/2` AC - - - - (iv)

Now, from equation (i) and equation (iii), we get

EF||HG

And, from equation (ii) and equation (iv), we get

EF = HG

Now, since one pair of opposite sides are parallel and equal in quadrilateral EFGH, therefore, EFGH is a parallelogram.

And we know that diagonals of a parallelogram bisect each other.

Thus, diagonals HF and EG of parallelogram EFGH bisect each other.

Therefore, line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Proved

Quadrilaterals Class 9 math NCERT Exercise 8.2 Question (7) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC

(ii) MD `_|_` AC

(iii) CM = MA = `1/2` AB

Solution

Let, ABC is the given right angled triangle.

In which ∠C = 90^o

And, M is the mid-point of hypotenuse AB

MD||BC

Thus, to prove that

(i) D is the mid-point of AC

In triangle ABC

M is the mid-point of side AB

And MD ||BC

[As given]

Now, we know from the Converse of Mid-Point Theorem that The line drawn through mid-point of one side of a triangle parallel to another side bisects the third side.

Therefore, MD bisects the third side AC of the given triangle.

That is CD = DA

Thus, D is the mid-point of AC Proved

To prove (ii) MD `_|_` AC

AS given, BC||MD

And a transversal AC is intersecting these two parallel lines BC and MD

Now, we know from the one of the theorem of parallel lines which says that If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

Here, ∠BCD and MDA are a pair of corresponding angles, and hence are equal.

That is ∠MDA = ∠BCD

⇒ ∠MDA = 90^o

[Because ∠C is 90^o as given in question]

Therefore, `MD`_|_`AC Proved

To prove (iii) CM = MA = `1/2`AB

Let join points M and C

In triangles CDM and DAM,

CD = DA

[Because D is the mid-point of AC as has already been proved in section (i)]

∠CDM = ∠MDA = 90^o

[Because MD is perpendicular to AC as has already been proved in section (ii)]

And, MD is common side in both of the triangles

Therefore, by SAS (Side Angle Side) congruency criterion

ΔCDM ≅ ΔDAM

Now, from CPCT, we know that, corresponding parts of congruent triangles are equal,

Therefore, CM = MA - - - - (i)

Now, since M is the mid-point of AB (As given in the question)

Therefore, MA = `1/2`AB - - - (ii)

Now, from equation (i) and equation (ii), we get

CM = MA = `1/2`AB Proved

Summary

(1) Sum of the angles of a quadrilateral is 360^o

(2) A diagonal of a parallelogram divides it into two congruent triangles.

(3) In a parallelogram, opposite sides are equal.

(4) In a parallelogram, opposite angles are equal.

(5) In a parallelogram, diagonals bisect each other

(6) A quadrilateral is a parallelogram, if opposite sides are equal

(7) A quadrilateral is a parallelogram, if opposite angles are equal

(8) A quadrilateral is a parallelogram, if diagonals bisect each other

(9) A quadrilateral is a parallelogram, if a pair of opposite sides is equal and parallel

(10) Diagonals of a rectangle bisect each other and are equal and vice-versa

(11) Diagonals of a rhombus bisect each other at right angle

(12) Diagonals of a square bisect each other at right angles are equal

(13) The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.

(14) A line through the mid-point of a side of a triangle parallel to another side bisects the third side.

(15) The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram.

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