Quadrilaterals: class 9 math: 9 Math

NCERT Exercise 8.1:Part-3: 9th math

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (9) In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution

Given, ABCD is a parallelogram and BD is its diagonal.

Two points P and Q lie on diagonal BD in such a way that

DP = BQ

Therefore, to prove

(i) ΔAPD ≅ ΔCQB

In ΔAPD and ΔCQB

AD = BC

[Because opposite sides of a parallelogram are equal. Here AD and BC are opposite sides of given parallelogram and hence are equal].

DP = BQ [As given in question].

Since ABCD is a parallelogram, hence AD||BC.

Here a transversal BD is intersecting the parallel lines AD and BC thus, alternate interior angles will be equal.

Therefore; ∠PDA = ∠CBQ.

Therefore, by SAS (Side Angle Side) congruency criterion,

Δ APD ≅ Δ CQB. Proved

(ii) AP =CQ.

Now PDA ≅ Δ CQA.

[As proved in equation (i) above]

Therefore, by CPCT we know that corresponding parts of congruent triangles are equal.

Therefore, AP = CQ Proved

(iii) Δ AQB ≅ Δ CPD

AB = CD

[Because opposite sides of a parallelogram are equal]

DP = BQ [As given in question]

Since ABCD is a parallelogram, hence AB||DC.

Here a transversal BD is intersecting the parallel lines AB and DC thus, alternate interior angles will be equal.

Thus, ∠CDP = ∠ABQ

Now, from SAS (Side Angle Side) congruency criterion,

ΔAQB ≅ ΔCPD Proved

(iv) AQ = CP

Since, ΔAQB ≅ ΔCPD

[As proved in the section (iii) above]

Thus, by CPCT we know that corresponding parts of congruent triangles are equal

Thus, AQ = CP Proved

(v) APCQ is a parallelogram

From section (ii) and section (iv) we have

AP = CQ and AQ = CP

Here opposite sides of APCQ are equal.

Therefore, APCQ is a parallelogram Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (10) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Solution

Given, ABCD is a parallelogram and BD is its diagonal.

And AP and CQ are two perpendiculars from vertices A and C on the diagonal BD.

Then to prove

(i) ΔAPB ≅ ΔCQD

In triangles APB and CQD,

AB = DC

[Because these are opposite sides of given parallelogram, and hence are equal]

∠APB = ∠CQD = 90^o

[As given in question AP and CQ are perpendicular to BD]

Here AB||DC

[Because ABCD is a parallelogram and we know that opposite sides of a parallelogram are parallel]

And a transversal BD is intersecting these parallel lines AB and CD

Thus, alternate interior angles will be equal.

Thus, ∠ABP = ∠CDQ

Now, from AAS (Angle Angle Side) congruency criterion

ΔAPB ≅ ΔCQD Proved

To prove (ii) AP = CQ

ΔAPB ≅ ΔCQD

[Proved in section (i)]

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

Thus, AP = CQ Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (11) In ΔABC and ΔDEF, AP = DE, AB||DE, BC = EF and BC||EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure).

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD||CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF

Solution

Given, ABC and DEF are two triangles in which

AB = DE, AB||DE

And, BC = EF and BC||EF

Vertices A, B and C are joined to vertices D, E and F respectively

Then to prove that

(i) Quadrilateral ABED is a parallelogram

In quadrilateral ABED

As given, AB|| ED

And AB = DE

Here, since one pair of opposite sides are parallel and equal in a quadrilateral.

Consequently, other opposite sides will also be parallel and equal.

Since, opposite sides are parallel and equal in the given quadrilateral

Therefore, ABED is a parallelogram Proved

To prove (ii) Quadrilateral BEFC is a parallelogram

In quadrilateral BEFC,

BC||EF and BC = EF

Since, one of the pair of opposite sides of the given quadrilateral are parallel and equal, consequently other pair of opposite sides will be parallel and equal.

Therefore, given quadrilateral BEFC is a parallelogram Proved

To prove (iii) AD||CF and AD = CF

In quadrilateral ACFD,

As proved in section (i) ABED is a parallelogram

Since ABED is a parallelogram

Therefore, AD||BE - - - - - (i)

And AD = BE - - - - - (ii)

And since, BEFC is a parallelogram as proved in section (ii)

Therefore, BE||CF - - - - - (iii)

And, BE = CF - - - - - (iv)

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

AD||CF

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

AD = CF

Thus, AD||CF and AD = CF Proved

To prove that (iv) quadrilateral ACFD is a parallelogram

As in section (iii) it has been proved that AD||CF and AD = CF

Now, since one pair of opposite sides are parallel and equal in quadrilateral ACFD, consequently other pair of opposite sides will be parallel and equal.

Thus, ACFD is a parallelogram Proved

To prove that (v) AC = DF

In section (iv) it has been proved that ACFD is a parallelogram.

And we know that, opposite sides of a parallelogram are parallel and equal.

Therefore, AC = DF Proved

To prove that (vi) ΔABC ≅ ΔDEF

AB = DE and BC = EF [As given in question]

And in section (v) it has been proved that AC = DF

Thus, by SSS (Side Side Side) congruency criterion,

ΔABC ≅ ΔDEF Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (12) ABCD is a trapezium in which AB||CD and AD = BC (see figure).

Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC &cong: ΔBAD

(iv) Diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E]

Solution

Given, ABCD is a trapezium

In which AB||CD and AD = BC

Construction

According to hint, AB has been extended

And a line CE parallel to DA has been drawn.

Then to prove that

(i) ∠A = ∠B

Here, AD||CD [As per construction]

And AB||DC [As given in question]

Therefore, ADCE is a parallelogram.

Therefore, AD = CE - - - - - (i)

[Because opposite sides of a parallelogram are equal.]

And, AD = BC - - - - -(ii)

[As given in the question]

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

BC = CE - - - - - (iii)

Now we know that in a triangle angles opposite to equal sides are equal.

Therefore, ∠EBC = ∠CEB - - - - - (iv)

Here a transversal AE is intersecting two parallel lines AD and CE. And we know that sum of angles of the same side of a transversal is equal to 180^o

Therefore, ∠A + ∠CEB = 180^o

⇒ ∠A + ∠EBC = 180^o - - - - - (v)

[Since from equation (iv) ∠EBC = ∠CEB]

Now, since ∠CBA and ∠EBC are forming a linear pair of anlges

Therefore, ∠CBA + ∠ EBC = 180^o - - - - (vi)

Now, from equation (v) and equation (vi), we get

∠A = ∠CBA

⇒ ∠A = ∠B Proved

To prove that (ii) ∠C = ∠D

Here, AB||CD [As given in the question]

And a transversal is intersecting parallel lines AB and CD.

And we know that sum of angles of the same side of a transversal which is intersecting two lines is equal to 180^o

Therefore, ∠A + ∠D = 180^o - - - - - (vii)

And again since ∠B and ∠C lie on the same side of the transversal BC.

Therefore, ∠B + ∠C = 180^o

In section (i) it has been proved that ∠A = ∠B.

Therefore, by replacing ∠B = ∠A in the above expression, we get

∠A + ∠C = 180^o - - - - - (viii)

Now, by equation (vii) and equation (viii), we get

∠A + ∠C = ∠A + ∠D

⇒ ∠C = ∠D Proved

To prove that (iii) ΔABC ≅ ΔBAD

Let join A and C and B and D in the figure given in question

Now, in ΔABC and ΔBAD

BC = AD [As given in question]

∠A = ∠B [As proved in section (i)]

And side AB is common in both of the triangles.

Therefore by SAS (Side Angle Side) congruency criterion

ΔABC ≅ ΔBAD Proved

To prove that (iv) Diagonal AC = diagonal BD

ΔABC ≅ ΔBAD

[Already proved in section (iii)]

Therefore, by CPCT, we know that corresponding parts of congruent triangles are equal.

Therefore, AC = BD

i.e. diagonal AC = diagonal BD Proved

Back to 9-math-home

9th-math (Hindi)

Reference: