# 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**

**Calculate:**

**(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.**

**Answer**

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**

**Answer**In △AEG,

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**

**Answer**

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 **

**Answer**

**(a)**

**(b)**

**(c)**

**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)**

**12. ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that:**

**(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.**

**Answer**