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Let, EC = x cm
It is given that DE || BC
By using basic proportionality theorem, we obtain
.png)
x = 2
EC = 2 cm.
(ii) Let AD = x cm.
It is given that DE || BC.
By using basic proportionality theorem, we obtain
x = 2.4
AD = 2.4cm.
2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF II QR:
(i) PE = 3.9 cm, EQ = 3cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Hence,
.png)
Therefore, EF is not parallel to QR.
(ii)Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
.png)
Therefore,EF is parallel to QR.
(iii)Given: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
And ER = PR – PF = 2.56 – 0.36 = 2.20 cm
.png)
.png)
Hence,
Therefore, EF is parallel to QR.
3. In figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
.png)
By using Basic Proportionality theorem
And in ∆ ACD, LN || CD
.png)
......i
Similalry, in ∆ ACD, LN || CD
.png)
.......ii
By using Basic Proportionality theorem
From eq. (i) and (ii), we have
4. In figure, DE || AC and DF || AE. Prove that BF/GE = BE/EC .
[Basic Proportionality theorem] ……….(i)
And in ∆ BEA, DF || AE
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
5. In figure, DE || OQ and DF || OR. Show that EF || QR.
.png)
[Basic Proportionality theorem] ……….(i)
And in ∆ POR, DF || OR
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
Therefore, EF || QR [By the converse of Basic Proportionality Theorem]
6. In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
And in ∆ POQ, AB || PQ
[Basic Proportionality theorem] ……….(i)
And in ∆ OPR, AC || PR
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
Therefore, BC || QR (By the converse of Basic Proportionality Theorem)
7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
.png)
Consider the given figure in which l is a line drawn through the mid-point P of line segment AB meeting AC at Q, such that PQ || BC
By using Basic Proportionality theorem, we obtain,
(P is the midpoint of AB ∴ AP = PB)
⇒ AQ = QC
Or, Q is the mid-point of AC.
8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
.png)
Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
i.e., AP = PB and AQ = QC
It can be observed that
.png)
and
Therefore,
Hence, by using basic proportionality theorem, we obtain
PQ || BC.
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO .
Draw a line EF through point O, such that EF || CD
In ΔADC, EO || CD
By using basic proportionality theorem, we obtain
...i
In ΔABD, OE || AB
So, by using basic proportionality theorem, we obtain
⇒
...ii
From eq. (i) and (ii), we get
⇒
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO Such that ABCD is a trapezium.
Quadrilateral ABCD is a trapezium.
Construction: Through O, draw OE || AB meeting AD at E.
In ∆ ADB, we have OE || AB [By construction] By Basic Proportionality theorem
.....i
However, it is given that
.....ii
From eq. (i) and (ii), we get
⇒ EO || DC [By the converse of basic proportionality theorem]
⇒ AB || OE || DC
⇒ AB || CD
∴ ABCD is a trapezium.
In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]
∴
..... [Corresponding sides are proportional]
Hence,
4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR .....(i)
Given,
Using (i), we get,
In ΔPQS & In ΔTQR
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴By AA-criterion of similarity, we have
∴ ΔABE ~ ΔCFB (By AA similarity criterion)
9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
