ICSE Solutions for Selina Concise Chapter 12 Mid Point and its Converse (including Intercept Theorem) Class 9 Maths
Exercise 12(A)
1. In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC in N. Find the lengths of AN and MN if Bc = 7 cm and Ac = 5 cm.Answer
The triangle is shown below,
MN = (1/2)BC = (1/2)×7 = 3.5cm
and AN = (1/2)AC = (1/2)×5 = 2.5 cm
2. Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
Answer
The figure is shown below,
Let ABCD be a rectangle where P,Q,R,S are the midpoint of AB, BC, CD, DA. We need to show that PQRS is a rhombus for help we draw two diagonal BD and AC as shown in figure .
where BD = AC (Since diagonal of rectangle are equal)
Proof :
From △ABD and △BCD
PS = (1/2)BD = QR and PS ॥ BD ॥ QR
2PS = 2QR = BD and PS ॥ QR ...(1)
Similarly 2PQ = 2SR = AC and PQ||SR ...(2)
From (1) and (2) we get
PQ = QR = RS = PS
Therefore, PQRS is a rhombus.
Hence, proved
Answer
The figure is shown below
Given that ABC is an isosceles triangle where AB=AC.
Since D,E,F are midpoint of AB,BC,CA therefore
2DE = AC and 2EF = AB this means DE = EF
Therefore DEF is an isosceles triangle an DE = EF.
Hence, proved
Answer
Here from triangle ABD P is the midpoint of AD and PR||AB, therefore Q is the midpoint of BD
Similarly R is the midpoint of BC as PR||CD||AB
From triangle ABD,
2PQ=AB ...(1)
From triangle BCD,
2QR=CD ...(2)
Now (1)+(2) we get,
2(PQ+QR) = AB+CD
⇒ PR = (1/2)(AB + CD)
Hence, proved
(i) MN, if AB = 11 cm and DC = 8 cm.
(ii) AB, if Dc = 20 cm and MN = 27 cm.
(iii) DC, if MN = 15 cm and AB = 23 cm.
Answer
Let us draw a diagonal AC as shown in the figure below,
(i) Given that,
AB = 11 cm, CD = 8 cm
From triangle ABC,
ON = (1/2)AB = (1/2)×11 = 5.5cm
From triangle ACD,
Om = (1/2)CD = (1/2)×8 = 4 cm
Hence MN = OM + ON = (4+5.5) = 9.5cm
(ii) Given that,
CD = 20 cm , MN = 27 cm
From triangle ACD,
OM = (1/2)CD = (1/2)×20 = 10 cm
Therefore ON = 27-10 = 17cm
From triangle ABC,
AB = 2 × ON = 2× 17 = 34 cm
(iii) Given that,
AB = 23cm, MN = 15cm
From triangle ABC,
ON = (1/2)AB = (1/2)×23 = 11.5 cm
Therefore, OM = 15 - 11.5 = 3.5cm
From triangle ACD,
CD = 2×OM = 2×3.5 = 7cm
6. The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is rectangle.
Answer
The figure is shown below
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
Therefore PQRS is a parallelogram.
Now PQ॥ AC, therefore ∠AOD = ∠PXO = 90° [Corresponding angle]
Again BD ॥ RQ, therefore ∠PXO = ∠RQX = 90° [Corresponding angle]
Answer
The required figure is shown below
From figure,
BL = DM and BL||DM and BLMD is a parallelogram, therefore BM||DL
From triangle ABY,
L is the midpoint of AB and XL||BY, therefore x is the midpoint of AY.ie AX = XY ...(1)
Similarly for triangle CDX
CY = XY ...(2)
From (1) and (2)
AX = XY = CY and AC = AX + XY + CY
Hence, proved
Answer
Given that AD=BC ...(1)
From the figure,
For triangle ADC and triangle ABD
2GH = AD and 2EF = AD, therefore 2GH = 2EF = AD ...(2)
For triangle BCD and triangle ABC
2GF = BC and 2EH = BC, therefore 2GF = 2EH = BC ...(3)
From (1),(2),(3) we get,
2GH = 2EF = 2GF = 2EH
GH = EF = GF = EH
Therefore EFGH is a rhombus.
Hence, proved
Prove that
(i) R is the midpoint of BC
(ii) PR = (1/2)DB
Answer
For help we draw the diagonal BD as shown below
The diagonal AC and BD cuts at point x.
We know that the diagonal of a parallelogram intersects equally each other. Therefore
Given,
CQ = (1/4)AC
⇒ CQ = (1/4)× 2CX
⇒ CQ = (1/2)CX
Therefore Q is the midpoint of CX.
(i) For triangle CDX PQ॥ DX or PR॥ BD.
(ii) For triangle BCD
As P and R are the midpoint of CD and BC, therefore PR= (1/2)DB
Answer
The required figure is shown below
For triangle ABC and OBC
2DE = BC and 2PQ = BC, therefore DE=PQ ...(1)
For triangle ABO and ACO
2PD = AO and 2FQ = AO, therefore PD = FQ ...(2)
From (1),(2) we get that PQFD is a parallelogram.
Hence, proved
(i)AP=2AR
(ii)BC=4QR
Answer
The required figure is shown below
Therefore Q is the midpoint of AB and R is the midpoint of AP
(i) Therefore AP = 2AR
(ii) Here we increase QR so that it cuts AC at S as shown in the figure.
(iii) From triangle PQR and triangle ARS
∠PQR = ∠ARS (Opposite angle )
PR = AR
PQ = AS [PQ = AS = (1/2)AC]
△PQR ≅ △ARS (SAS Postulate)
Therefore QR = RS
Now
BC = 2QS
⇒ BC = 2×2QR
⇒ BC = 4QR
Hence, proved
(i) Point P bisects BE,
(ii) PQ is parallel to AB.
Answer
The required figure is shown below
(i) From △PED and △ABP
PD = AP [P is the midpoint of AD]
∠DPE = ∠APB [Opposite angle]
∴ ΔPED ≅ ΔABP [ASA postulate]
∴ EP = BP
(ii) For triangle ECB PQ||CE
Again, CE||AB
Therefore, PQ||AB
Hence, proved
Prove that: AC = 3AF
Answer
The required figure is shown below
For help we draw a line DG||BF
Now from triangle ADG, DG||BF and E is the midpoint of AD
Therefore, F is the midpoint of AG, i.e. AF = GF ...(1)
From triangle BCF, DG||BF and D is the midpoint of BC
Therefore G is the midpoint of CF, i.e. GF = CF …(2)
AC = AF + GF + CF
⇒ AC = 3AF (From 1 and 2)
Hence, proved
(i) Prove that BDFE is parallelogram
(ii) Find AB, if EF = 4.8 cm.
Answer
The required figure is shown below
(i) Since F is the midpoint and EF ॥ AB.
Therefore E is the midpoint of BC
So,
Since, D and F are the midpoint of AB and AC
Therefore,
So DF = (1/2)BC and DB = (1/2)AB ...(2)
From (1), (2) we get
BE = DF and BD = EF
Hence, BDEF is a parallelogram.
AB = 2EF
= 2×4.8
= 9.6cm
Answer
AD is the median of BC
⇒ D is the mid-point of BC.
Given that DE P BA
By the Converse of the Mid - point theorem,
⇒ DE bisects AC
⇒ E is the mid- point of AC
⇒ BE is the median of AC, that is BE is also a median.
Answer
Construction: Draw DY ॥ BQ
In ΔBCQ and ΔDCY,
ã„¥BCQ = ã„¥DCY (Common)
ã„¥BQC = ã„¥DYC (Corresponding angles)
So, ΔBCQ ～ΔDCY (AA Similarity criterion)
⇒ BQ/DY = BC/DC = CQ/CY (Corresponding sides are proportional)
⇒ BQ/DY = 2CD/CD (D is the mid - point of BC)
⇒ BQ/DY = 2 ...(i)
Similarly, ΔAEQ ～ ΔADY
⇒ EQ/DY = AE/ED = 1/2 (E is the mid-point of AD)
that is EQ/DY = 1/2 ...(ii)
Dividing (i) by (ii), we get
⇒ BQ/EQ = 4
⇒ BE + EQ = 4EQ
⇒ BE = 3EQ
⇒ BE/EQ = 3/1
Answer
Mis the mid- point of AB and Nis the mid- Point of DE.
⇒ MN = (1/2)EF (Mid-point theorem)
⇒ EF = 2MN ...(i)
In ΔABC,
M is the mid - point of AB and N is the mid- point of BC
⇒ MN = (1/2)AC (Mid - point theorem)
⇒ AC = 2MN ...(ii)
From (i) and (ii), we get
⇒ EF = AC
Exercise 12(B)
1. Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.
Answer
According to equal intercept theorem since CD=DE
Therefore AB=BC and EF=GF
(i) BC=AB=7.2cm
(ii) GE=EF+GF = 2EF = 8cm
Since B,D,F are the midpoint and AE||BF||CG
Therefore AE=2BD and CG=2DF
(iii) AE = 28D = 2×4.1 = 8.2
(iv) DF = (1/2) CG = (1/2)×11 = 5.5cm
(i) AQ||BS
(ii) DS = 3 Rs.
Answer
Given that AD = AP = PB as 2AD = AB and p is the midpoint of AB
(i) From triangle DPR, A and Q are the midpoint of DP and DR.
Therefore, AQ||PR
Since, PR||BS ,hence AQ||BS
(ii) From triangle ABC, P is the midpoint and PR||BS
Therefore R is the midpoint of BC
From △BRS and △QRC
∠BRS = ∠QRC
BR = RC
∠RBS = ∠RCQ
∴ △BRS ≅ △QRC
∴ QR = RS
⇒ DS = DQ + QR + RS = QR + QR + RS = 3RS
(i) 3DF = EF
(ii) 4CR = AB.
Answer
Consider the figure:
(i) Again from the triangle AEF we have AE ॥ PD ॥ CR and AP = (1/3)AE
Therefore DF = (1/3)EF or we can say that 3DF = EF.
(ii) From the triangle PED we have PD ॥ CR and C is the midpoint of PE therefore CR = (1/2)PD
Now,
PD = (1/2)AB
⇒ (1/2)PD = (1/4)AB
⇒ CR = (1/4)AB
⇒ 4CR = AB
Hence, it is shown.
(i) M, A and N are collinear.
(ii) A is the mid-point of MN
Answer
The figure is shown below
(i) From triangle BPC and triangle APN
∠BPC = ∠APN [Opposite angle]
BP = AP
PC = PN
∴ Δ BPC ≅ ΔAPN [SAS postulate]
∴ ∠PBC =∠PAN ...(1)
And BC = AN ...(3)
Similarly ∠QCB = ∠QAN ...(2)
And BC = AM ...(4)
Now
∠ABC + ∠ACB + ∠BAC = 180°
∠PAN + ∠QAM + ∠BAC = 180° [(1), (2) we get]
Therefore M, A, N are collinear
(ii) From (3) and (4) MA = NA
Hence, A is the midpoint of MN
Answer
The figure is shown below
From the figure EF||AB and E is the midpoint of BC.
Therefore, F is the midpoint of AC.
Here, EF||BD, EF=BD as D is the midpoint of AB
BE||DF, BE=DF as E is the midpoint of BC.
Therefore, BEFD is a parallelogram.
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
Answer
The figure is shown below
BE = FC
∠EHB = ∠FHC [Opposite angle]
∠HBE = ∠HFC
∴ △HEB ≅ △FHC
∴ EH = CH, BH = FH
For triangle ECD, F and H are the midpoint of CD and EC.
Therefore HF॥DE and HF = (1/2) DE ...(2)
(1),(2) we get, HF = EG and HF||EG
Similarly, we can show that EH = GF and EH||GF
Therefore, GEHF is a parallelogram.
Answer
The figure is shown below
For triangle AEG
D is the midpoint of AE and DF||EG||BC
Therefore, F is the midpoint of AG.
AF = GF ...(1)
Again DF||EG||BC, DE = BE, therefore GF = GC ...(2)
(1),(2) we get AF = GF = GC.
Similarly,
Since GN||FM||AB and AF = GF, therefore BM = MN = NC
Hence, proved.
Answer
The figure is shown below
Since M and N are the midpoint of AB and AC, MN||BC
According to intercept theorem,
Since MN||BC and AM = BM,
Therefore AX=DX.
Hence, proved
Answer
The figure is shown below
Diagonal AC and BD intersect at point O. We need to show that AC and BD intersect at right angle.
Proof :
PQ ॥ AC, therefore ∠AOD = ∠PXO [Corresponding angle] ...(1)
Again BD॥RQ, therefore ∠PXO = ∠RQX = 90° [corresponding angle and angle of rectangle]...(2)
From (1) and (2) we get,
∠AOD = 90°
Similarly ∠AOB = ∠BOC = ∠DOC = 90°
Hence, proved
Answer
The figure is shown below
Therefore F is the midpoint of BC and 2 DE = BC or DE = BF
Again,
Hence BDEF is a parallelogram.
Now,
BD = EF = (1/2)AB = (1/2)×16 = 8cm
BF = DE = (1/2)BC = (1/2)×18 = 9cm
Therefore, perimeter of BDEF = 2 (BF + EF) = 2(9+8) = 34 cm
Answer
Given AD and CE are medians and DF || CE.
We know that from the midpoint theorem, if two lines are parallel and the starting point of segment is at the midpoint on one side, then the other point meets at the midpoint of the other side.
Consider triangle BEC. Given DF || CE and D is midpoint of BC. So F must be the midpoint of BE.
So, FB = (1/2)BE but BE = (1/2)AB
Substitute value of BE in first equation, we get
FB = (1/4)AB
Hence, Proved
Prove that:
(i) BP = 2AD
(ii) O is the mid-point of AP.
Answer
Given,
ABCD is parallelogram, so AD = BC, AB = CD.
Consider triangle APB, given EC is parallel to AP and E is midpoint of side AB. So by midpoint theorem, C has to be the midpoint of BP.
So BP = 2BC, but BC = AD as ABCD is a parallelogram.
Hence, BP = 2AD
Consider triangle APB,
AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.
Hence, Proved
Answer
Consider trapezium ABCD.
Given E and F are midpoints on sides AD and BC, respectively.
We know that AB = GH = IJ
From midpoint theorem,
Consider LHS,
AB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG
So, AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF
AB + CD = 2EF
Hence, Proved
Answer
Given ΔABC,
AD is the median. So D is the midpoint of side BC.
Given DE || AB. By the midpoint theorem, E has to be midpoint of AC.
So line joining the vertex and midpoint of the opposite side is always known as median. So BE is also median of Δ ABC.
Answer
Now in right - angles △BOC,
