Frank Solutions for Class 9 Maths Chapter 15 Mid – point and Intercept Theorems act as an important tool in preparing for examinations. The solutions are prepared in accordance with the current ICSE board keeping in mind the understanding capacity of students. Practising solutions on a daily basis help students to achieve desired scores along with speeding up their problem solving skills. Frank Solutions are available in PDF to encourage concept based learning among students. Access Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems, from the available links below.
Chapter 15 has problems relying on mid-points of a triangle. Mid-point theorem of a triangle states that if the two triangle sides’ midpoints are connected then the resulting line segment is parallel to the third triangle side. Students can cross check their answers and clear doubts which appear while solving textbook problems.
Frank Solutions for Class 9 Maths Chapter 15 Mid – point and Intercept Theorems Download PDF
Access Frank Solutions for Class 9 Maths Chapter 15 Mid – point and Intercept Theorems
1. In â–³ABC, D is the mid-point of AB and E is the mid-point of BC
Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 720
Solution:
In â–³ABC,
D and E are the mid-points of AB and BC respectively
Hence, by mid-point theorem DE || AC and DE = (1/2) AC
(i) DE = (1/2) AC = (1/2) x 8.6
We get,
= 4.3 cm
(ii) ∠DEB = ∠C = 720 (Corresponding angles are equal, since DE || AC)
2. In â–³ABC, AB = 12 cm and AC = 9 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC in N, what is the length of MN?
Solution:
MN || AC and M is the mid-point of AB
Hence, N is the mid-point of BC
Therefore, MN = (1/2) AC
= (9/2) cm
We get,
= 4.5 cm
3. (a) In â–³ABC, D, E, F are the mid-points of BC, CA and AB respectively. Find FE, if BC = 14 cm
(b) In â–³ABC, D, E, F are the mid-points of BC, CA and AB respectively. Find DE, if AB = 8 cm
(c) In △ABC, D, E, F are the mid-points of BC, CA and AB respectively. Fine ∠FDB if ∠ACB = 1150
Solution:
(a)
F is the mid-point of AB and E is the mid-point of AC
Hence,
FE = (1/2) BC ………… (Mid-point Theorem)
= (1/2) x 14
We get,
= 7 cm
(b)
In â–³ABC,
D is the mid-point of BC and E is the mid-point of AC
Hence,
DE = (1/2) AB …….. (Mid-point Theorem)
= (1/2) x 8
We get,
= 4 cm
(c)
In â–³ABC,
FD || AC
Hence,
∠FDB = ∠ACB = 1150 …….(Corresponding angles are equal)
4. In parallelogram PQRS, L is mid-point of side SR and SN is drawn parallel to LQ which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ
Solution:
In â–³NSR,
MQ = (1/2) SR
But L is the mid-point of SR and SR = PQ (Sides of parallelogram)
So, it can be written as,
MQ = (1/2) PQ
MQ = PM = LS= LR
Hence, M is the mid-point of PQ
5. In â–³ABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that:
(a) QAP is a straight line
(b) A is the mid-point of PQ
Solution:
Since, BE and CF are the medians,
F is the mid-point of AB and E is the mid-point of AC
We know that the line joining the mid-points of any two sides is parallel and half of the third side
We have,
In â–³ACQ,
EF|| AQ and EF = (1/2) AQ ………(1)
In â–³ABP,
EF || AP and EF = (1/2) AP ……….(2)
(a) From (1) and (2)
We get,
AP || AQ (both are parallel to EF)
As AP and AQ are parallel and have a common point A
This is possible only if QAP is a straight line
Thus, proved
(b) From (1) and (2),
EF = (1/2) AQ and EF = (1/2) AP
⇒ (1/2) AQ = (1/2) AP
⇒ AQ = AP
Therefore, A is the mid-point of QP
6. Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus
Solution:
In the given rectangle ABCD,
Join AC and BD
In â–³ABC, P and Q are the mid-points of AB and BC respectively
PQ = (1/2) AC …….. (1) and PQ ||AC
In â–³BDC, R and Q are the mid-points of CD and BC respectively
QR = (1/2) BD ……..(2) and QR || BD
But AC and BD are diagonals of the rectangle
From equations (1) and (2)
PQ = QR
Similarly,
QR = RS and RS = SP
And,
RS || AC and SP || BD
Therefore, PQ = QR = RS = SP
Hence, PQRS is a rhombus
7. D, E and F are the mid-points of the sides AB, BC and CA of an isosceles â–³ABC in which AB = BC. Prove that â–³DEF is also isosceles.
Solution:
E and F are mid-points of BC and AC
Hence, EF = (1/2) AB …….(1)
D and F are the mid-points of AB and AC
Hence, DF = (1/2) BC ……..(2)
But AB = BC
From (1) and (2)
EF = DF
Thus, â–³DEF is an isosceles triangle
8. The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
Solution:
P and Q are the mid-points of AB and BC
Hence,
PQ || AC and PQ = (1/2) AC ……. (i)
S and R are the mid-points of AD and DC
Hence,
SR || AC and SR = (1/2) AC …….(ii)
From (i) and (ii),
PQ || SR and PQ = SR
Therefore, PQRS is a parallelogram
Further AC and BC intersect at right angles
∴ SP || BD and BD ⊥ AC
∴ SP ⊥ AC
⇒ SP ⊥ SR
⇒ ∠RSP = 900
∴ ∠RSP = ∠SRQ = ∠RQP = ∠SPQ = 900
Hence, PQRS is a rectangle
9. In a right angled triangle ABC. ∠ABC = 900 and D is the mid-point of AC. Prove that BD = (1/2) AC
Solution:
Draw line segment DE ||CB, which meets AB at point E
Now,
DE || CB and AB is the transversal,
∴∠AED = ∠ABC …. (corresponding angles)
∠ABC = 900 (given)
⇒ ∠AED = 900
Also,
Since D is the mid-point of AC and DE || CB,
DE bisects side AB,
Hence,
AE = BE …..(i)
In â–³AED and â–³BED,
∠AED = ∠BED …..(Each 900)
AE = BE ……[From (i)]
DE = DE ….(Common)
Therefore, △AED ≅ △BED ….(By SAS)
⇒ AD = BD …..(C.P.C.T.C)
⇒ BD = (1/2) AC
Hence, proved
10. In a parallelogram ABCD, E and F are the mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively. Prove that:
(a) △GEA ≅ △GFD
(b) △HEB ≅ △HFC
(c) EGFH is a parallelogram
Solution:
Since ABCD is a parallelogram,
AB = CD and AD = BC
Now,
E and F are the mid-points of AB and CD respectively,
Hence,
AE = EB = DF = FC ……….(1)
(a) In â–³GEA and â–³GFD,
AE = DF …… [From (1)]
∠AGE = ∠DGF ……… (vertically opposite angles)
∠GAE = ∠GFD …….. (Alternate interior angles)
Therefore,
△GEA ≅ △GFD
(b) In â–³HEB and â–³HFC,
BE = FC …..[From (1)]
∠EHB = ∠FHC (vertically opposite angles)
∠HBE = ∠HFC (Alternate interior angles)
Therefore,
△HEB ≅ △HFC
(c) In quadrilateral AECF,
AE = CF …..[From (1)]
AE || CF ……(since AB || DC)
Hence,
AECF is a parallelogram
⇒ EC || AF or EH || GF …… (i)
In quadrilateral BFDE,
BE = DF ….[From (1)]
BE || DF ….(since AB || DC)
⇒ BEDF is a parallelogram
⇒ BF || ED or HF || EG ……(ii)
From equations (i) and (ii),
We get,
EGFH is a parallelogram
11. In â–³ABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that:
(a) Q, A and P are collinear
(b) A is the mid-point of PQ
Solution:
In â–³BDC and â–³ADQ,
CD = DQ … (given)
∠BDC = ∠ADQ ……(vertically opposite angles)
BD = AD ……(D is the mid-point of AB)
Therefore,
△BDC ≅ △ADQ
⇒ ∠DBC = ∠DAQ (cpct)…….(i)
And, BC = AQ (cpct) …….(ii)
Similarly,
We can prove
△CEB ≅ △ AEP
⇒ ∠ECB = ∠EAP (cpct)……..(iii)
And, BC = AP (cpct)……..(iv)
(a) In â–³ABC,
∠ABC + ∠ACB + ∠BAC = 1800
⇒ ∠DBC + ∠ECB + ∠BAC = 1800
⇒ ∠DAQ + ∠EAP + ∠BAC = 1800 [From (i) and (iii)]
⇒ Q, A, P are collinear
(b) From (ii) and (iv),
AQ = AP
Therefore, A is the mid-point of PQ
12. In â–³ABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC
Solution:
In â–³AEG,
D is the mid-point of AE and DF || EG
Hence,
F is the mid-point of AG
AF = FG ……….. (1)
In â–³ABC,
DF || EG|| BC
DE = BE
Hence,
GF = GC …………. (2)
From (1) and (2) we get,
AF = FG = GC
Similarly, since GN|| FM|| AB
Thus, BM = MN = NC (proved)
13. In the given figure, the lines l, m and n are parallel to each other. D is the mid-point of CE. Find
(a) BC
(b) EF
(c) CG
(d) BD
Solution:
According to equal intercept theorem,
Since CD = DE
AB = BC ………(i)
EF = GF ……….(ii)
(a) BC = AB = 6 cm ….. [From (i)]
(b) EG = EF + FG
EG = 2EF …… [From (ii)]
9 = 2EF
EF = (9/2)
EF = 4.5 cm
(c) CG = 2DF
CG = 2 x 4.2
CG = 8.4 cm
(d) AE = 2BD
BD = (1/2) AE
BD = (1/2) x 12
We get,
BD = 6 cm
14. The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the mid-points of quadrilateral ABCD is a rectangle.
Solution:
The figure is as shown below
Let ABCD be a quadrilateral where P, Q, R, S are the mid-points of sides AB, BC, CD, DA. Diagonals AC and BD intersect at point ‘O’.
We need to prove that PQRS is a rectangle
Proof:
In â–³ABC and â–³ADC,
2PQ = AC and PQ || AC …….. (1)
2RS = AC and RS || AC …….. (2)
From (1) and (2)
We get,
PQ = RS and PQ || RS
Similarly, we can show that
PS = RQ and PS || RQ
Hence,
PQRS is a parallelogram
PQ || AC
Therefore, ∠AOD = ∠PXO = 900 …. [Corresponding angles]
Again BD || RQ
Therefore, ∠PXO = ∠RQX = 900 …[Corresponding angles]
Similarly,
∠QRS = ∠RSP = ∠SPQ = 900
Hence,
PQRS is a rectangle
15. In â–³ABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
Solution:
Note: The given question is incomplete
According to the question given, F could be any point on BC as shown below
So, this makes it impossible to prove DP = DE
Since P too would shift as F shift because P too would be any point on DE as F is
Note: If we are given F to be the mid-point of BC, the result can be proved
Here,
D and E are the mid-points of AB and AC respectively
DE || BC and DE = (1/2) BC
But F is the mid-point of BC
BF = FC = (1/2) BC = DE
Since D is the mid-point of AB, and DP || BF
Since P is the mid-point of AF and E is the mid-point of AC,
PE = (1/2) FC
Also,
D and P are the mid-points of AB and AF respectively
⇒ DP = (1/2) BF = (1/2) FC = PE …….. (since BF = FC)
⇒ DP = PE
Hence, proved
