Frank Solutions for Chapter 15 Mid-point and Intercept Theorems Class 9 Mathematics ICSE
Exercise 15.1
1. In △ABC, D is the mid-point of AB and E is the mid-point of BC
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 72°
Answer
In △ABC,
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) ×8.6
We get,
= 4.3 cm
(ii) ∠DEB = ∠C = 72° (Corresponding angles are equal, since DE || AC)
2. In △ABC, 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?
Answer
Hence, N is the mid-point of BC
Therefore, MN = (1/2) AC
= (9/2) cm
We get,
= 4.5 cm
3. (a) In △ABC, D, E, F are the mid-points of BC, CA and AB respectively. Find FE, if BC = 14 cm
(b) In △ABC, 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 = 115°
Answer
(a)
Hence,
FE = (1/2) BC …(Mid-point Theorem)
= (1/2) x 14
We get,
= 7 cm
(b)
In △ABC,
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) ×8
We get,
= 4 cm
(c)
In △ABC,
FD || AC
Hence,
∠FDB = ∠ACB = 115° …(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.
In △NSR,
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 △ABC, 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
Answer
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 △ACQ,
EF|| AQ and EF = (1/2) AQ …(1)
In △ABP,
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.
Answer
Join AC and BD
In △ABC, P and Q are the mid-points of AB and BC respectively
PQ = (1/2) AC …(1) and PQ ||AC
In △BDC, 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 △ABC in which AB = BC. Prove that △DEF is also isosceles.
Answer
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, △DEF 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.
Answer
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 = 90°
∴ ∠RSP = ∠SRQ = ∠RQP = ∠SPQ = 90°
Hence, PQRS is a rectangle
9. In a right angled triangle ABC. ∠ABC = 90° and D is the mid-point of AC. Prove that BD = (1/2) AC
Answer
Now,
DE || CB and AB is the transversal,
∴ ∠AED = ∠ABC …(corresponding angles)
∠ABC = 90° (given)
⇒ ∠AED = 90°
Also,
Since D is the mid-point of AC and DE || CB,
DE bisects side AB,
Hence,
AE = BE …(i)
In △AED and △BED,
∠AED = ∠BED …(Each 90°)
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
Answer
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 △GEA and △GFD,
AE = DF ...[From (1)]
∠AGE = ∠DGF …(vertically opposite angles)
∠GAE = ∠GFD …(Alternate interior angles)
Therefore,
△GEA ≅ △GFD
(b) In △HEB and △HFC,
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 △ABC, 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
Answer
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 △ABC,
∠ABC + ∠ACB + ∠BAC = 180°
⇒ ∠DBC + ∠ECB + ∠BAC = 180°
⇒ ∠DAQ + ∠EAP + ∠BAC = 180° [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 △ABC, 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
D is the mid-point of AE and DF || EG
Hence,
F is the mid-point of AG
AF = FG …(1)
In △ABC,
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
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 = 2×4.2
⇒ CG = 8.4 cm
(d) AE = 2BD
⇒ BD = (1/2) AE
⇒ BD = (1/2)×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.
Answer
The figure is as shown below
We need to prove that PQRS is a rectangle
Proof:
In △ABC and △ADC,
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 = 90° …[Corresponding angles]
Again BD || RQ
Therefore, ∠PXO = ∠RQX = 90° …[Corresponding angles]
Similarly,
∠QRS = ∠RSP = ∠SPQ = 90°
Hence,
PQRS is a rectangle
15. In △ABC, 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.
Answer
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
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
Exercise 15.2
1. Prove that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer
2. If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
Answer
3. In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = 1/2.AC.
Answer
4. In parallelogram ABCD, P is the mid-point of DC. Q is point on AC such that CQ = 1/4 AC. PQ produced meets BC at R. Prove that
(i) R is the mid-point of BC, and
(ii) PR = ½.DB.
Answer
5. In a parallelogram ABCD, M is the mid-point of AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:
(i) Triangle AXM is congruent to triangle CYM, and
(ii) XMY is a straight line.
Answer
6. Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square of a square is also a square.
Answer
7. In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find :
(a) PQ, if AB = 12 cm and DC = 10 cm.
(b) AB, if DC = 8 cm and PQ = 9.5 cm
(c) DC, if AB = 20 cm and PQ = 14 cm
(b)
8. In AD is a median of side BC of △ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF : AC = 1 : 3.
Answer
Construction : Draw DS || BF, meeting AC at S.
9. M and N divide the side AB of △ABC into three equal parts. Line segments MP and NQ are both parallel to BC, and meet AC at P and Q respectively. Prove that P and Q divide AC into three equal parts.
Answer
10. △ABC is an isosceles triangle with AB = AC. D, E and F are the mid-points of BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.
Answer
11. ABCD is a parallelogram. E is the mid-point of CD and P is a point on AC such that PC = 1/4AC. EP produced meets at BC at F. Prove that:
(i) F is the mid-point of BC
(ii) ZEF = BD
Answer
(i)
(i) ∠EFG = 90°
(ii) The line drawn through G and parallel to FE and bisects DA.
Answer
(i)
(ii)
13. In △ABC, X is the mid-point of AB, and Y is the mid-point of AC, BY and CX are produced and meet the straight line through A parallel to BC at P and Q respectively. Prove AP = AQ.
Answer
14. In △ABC, D, E and F are the midpoints of AB, BC and AC.
(a) Show that AE and DF bisect each other.
(b) If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.
Answer
(a)
(b)
15. In the given figure, T is the midpoint of QR. Side PR of △PQR is extended to S such that R divides PS in the ratio 2 : 1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2: 1 and WR = ¼.PQ.
