Selina Concise Class 9 Maths Chapter 14 Rectilinear Figures Exercise 14C Solutions
EXERCISE – 14C
(1) E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD prove that AEFD is a parallelogram.
Solution :
Given,
E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD.
To prove : AEFD is a parallelogram.
Proof : AB = CD (opposite side of parallelogram)
AE + EB = DF + FC
AE + AE = DF + DF
2AE = 2DF
AE = DF
AE parallel DF (AE || DF)
∴ AEFD is a parallelogram.
(2) The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is rhombus.
Solution :
Given,
The diagonal BD of a parallelogram ABCD bisects angles B and D.
To prove : ABCD is a rhombus
Proof: In △ADB and △CDB,
∠y = ∠y
BD = BD
∠x = ∠x
∴ △ADB ≅ △CDB (by angle –side-angle (ASA))
=> AD = CD … (i)
BC = AB …..(ii)
AB = CD (iii) {by corresponding pair of congruent triangles is also congruent (CPCTC)}
From equation (i), (ii) and (iii),
AB = BC = CD = AD
Hence, ABCD is a rhombus.
(3) The alongside figure shows a parallelogram ABCD in which AB = EF = FC.
Prove that
(i) DE is parallel to FB
(ii) DE = FB
(iii) DEBF is a parallelogram.
Solution:
Given: ABCD is a parallelogram.
AE = EF = FC
To prove :
(i) DE || FB
(ii) DE = FB
(iii) DEBF is a parallelogram.
Proof :
In △AED and △CEB,
∠DAE = ∠BCF (Alternate interior angles)
AE = FC (opposite sides of parallelogram)
AD = BC (opposite sides of parallelogram)
∴ △AED ≅ △CFB (by side-angle-side (SAS)
DE = BF (by CPCTC)
∠AED = ∠BFC (by CPCTC)
Then,
DE parallel BF (DE || BF)
Hence, DEBF is a parallelogram.
(4) In the alongside diagram, ABCD is a parallelogram in which AP bisect angle AP bisects angle A and BQ bisects angle B. Prove that:
(i) AQ = BP
(ii) PQ = CD
(iii) ABPQ is a parallelogram.
Solution :
Given,
ABCD is a parallelogram.
and AP bisect ∠A and BQ bisect ∠B
To prove:
(i) AQ = BP
(ii) PQ = CD
(iii) ABPQ sq parallelogram.
Proof :
∠A + ∠B = 180° (co-interior angles)
2x + 2y = 180°
2 (x + y) = 180°
x + y = 180°/2
x + y = 90°
In △AOB,
Sum of all interior angles = 180°
∠OAB + ∠OBA + ∠AOB = 180°
x + y + ∠AOB = 180°
90° + ∠AOB = 180°
∠AOB = 180° – 90°
∠AOB = 90°
AP and BQ intersect each other at 90°
Hence, ABPQ is a rhombus.
Then,
AQ = BP
Also,
AB = QP …,. (i)
AB = CD ….. (ii)
From equation (i) and (ii),
QP = CD
If AQ = PB and AD parallel PB (AD || PB)
Hence ABPQ is parallelogram.
(5) In the given figure, ABCD is a parallelogram. Prove that : AB = 2B
Solution :
Given,
ABCD is a parallelogram.
AE bisect ∠A and BE bisect ∠B
To prove : AB = 2BC
Proof :
∠EAB = ∠AED = x (Alternate interior angle)
∠ABE = ∠BEC = y
(Alternate Interior angle)
In △ADE,
AD = DE …. (i) (because ∠DAE ∠DEA)
In △BCE,
BC = EC ….. (ii)
Now,
AB = CD
AB = DE + EC
AB = AD + BC
AB = BC + BC {From equation (i) and (ii)}
AB = 2BC [Hence proved]
Here is your solution of Selina Concise Class 9 Maths Chapter 14 Rectilinear Figures Exercise 14C
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