## Solid Geometry Chapter 2 Theorems On Transversal And Mid Points

**Chapter 2 Theorems On Transversal And Mid Points Theorems On Mid Points**

Let ABC be any triangle. D and E are the mid-points of its AB and AC sides respectively.

Then, DE || BC and DE = \(\frac{1}{2}\) BC.

Similarly, if F is the mid-point of BC, then DF || AC

and DF = \(\frac{1}{2}\) AC, EF || AB and EF = \(\frac{1}{2}\) AB

Therefore, the straight line obtained by joining the mid-points of any two sides of a triangle is parallel to its third side and is half of its length.

**Read and Learn More WBBSE Solutions For Class 9 Maths**

We shall now prove this theorem geometrically.

## Theorem 1. The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it in length.

**Given:** Let D and E be the mid-points of the sides AB and AC respectively of the triangle ABC.

Let us join D and E.

**To prove** DE || BC and DE = \(\frac{1}{2}\) BC.

**Construction:** Let DE be extended to F such that DE EF and let us join F, C.

**Proof:** In triangles, ΔADE and ΔCEF, AE = CE

[∵ E is the mid-point of AC. ]

<AED = <CEF [∵ opposite angles ] and DE = EF [by construction]

∴ ΔADE ≅ ΔCEF [ by the condition of S-A-S congruent ]

∴ AD = FC

or, BD = CF [since D is the mid-point of AB, AD = BD]

Again, DAEFCE, but these are alternative angles,

∴ DA || CF

or, BA || CF

or, BD || CF.

Thus, in the quadrilateral BDFC, BD || CF and BD = CF,

∴ BDFC is a parallelogram.

DF || BC and DF = BC

or, DE || BC and DE+ EF = BC

or, DE + DE = BC [∵ DE = EF]

or, 2DE = BC

or, DE = \(\frac{1}{2}\) BC

∴ DE = BC and DE = \(\frac{1}{2}\) BC. (Proved)

We shall now prove the converse of Theorem-1.

## Theorem 2. The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

**Given:** Let us draw DE parallel to BC through the mid-point D of AB of the ΔABC, which intersects AC at E.

**To prove : **AE = CE and DE = \(\frac{1}{2}\) BC.

**Construction: **ED is extended to F such that ED = DF.

Let us join B, F.

**Proof:** In triangles A ADE and A BDF, AD = BD [∵ D is the mid-point of AB ],

∠EDA = ∠FDB [… opposite angles] and ED=DF [by construction]

∴ AADE ≅ ABDF [ by the condition of S-A-S congruence]

∴ AE = BF [ similar sides of congruent triangles ]

∠AED= ∠BFD. But these are alternate angles.

∴ AE || FB or, EC || FB…… (1)

Also, given that DE || BC or, FE || BC…… (2)

From (1) and (2) we get, BCEF is a parallelogram.

FB = CE or, AE = CE [ FB = AE] and FE = BC

[∵ opposite sides of any parallelogram are equal ]

or, FD + DE = BC or, DE + DE = BC [∵ FD = DE ]

or, 2DE = BC or, DE= \(\frac{1}{2}\) BC

∴ AE = CE and DE = \(\frac{1}{2}\) BC. (Proved)

We shall now prove theorem-1 with the help of the application of theorem-2.

## Alternate proof of THEOREM-2

**Given:** Let D and E be the mid-points of the sides AB and AC respectively of the AABC and let us join D and B.

**To prove: **DE || BC and DE= \(\frac{1}{2}\) BC.

**Construction:** Let us draw EF through E, the mid-point of AC**, **parallel to AB, which intersects BC at F.

**Proof:** E is the mid-point of AC and EF || AB [by construction]

F is the mid-point of BC, i.e., BF = \(\frac{1}{2}\) BC and EF = \(\frac{1}{2}\) AB.

Again, EF = \(\frac{1}{2}\) AB = DB [∵ D is the mid-point of AB.]

Thus, in the quadrilateral DBFE, EF = DB and EF || DB.

∴ DBFE is a parallelogram.

∴ DE || BF or, DE || BC and DE = BF

or, DE =\(\frac{1}{2}\) BC.

∴ DE || BC and DE = \(\frac{1}{2}\) BC. (Proved)

Now, PR 10 cm,

∴ SR- cm 5 cm.

∴ the required length of QS = 5 cm.

**Alternative Method:**

We know that the median of any triangle drawn from the vertex forming the right-angle of any right-angled triangle to its hypotenuse is half of the hypotenuse.

QS = \(\frac{1}{2}\) PR = \(\frac{1}{2}\)x10 cm = 5 cm.

∴ the required length of QS = 5 cm

## Chapter 2 Theorems On Transversal And Mid Points Theorems On Transversal

You have already studied in your previous classes the properties of the angles obtained by the intersection of a straight line (which is generally called the transversal) with two or more two straight lines (parallel or unparallel). In this chapter, we shall prove geometrically one theorem on this transversal.

## Theorem 1. If a transversal intersects equal parts from three or more than three parallel straight lines, then they also intersect equal parts from another transversal.

**Given:** The parallel straight lines AB, CD and EF have intersected two equal parts GH and HI from the transversal PQ, i.e., GH = HI. The same parallel straight lines have also intersected A two parts LM and MN from another transversal RS.

**To prove **LM = MN.

**Construction: **Let us join G and N.

Also, let GN intersects CD at point T.

**Proof:** In triangle GIN, H is the mid-point of GI

[∵ GH = HI] and HT || IN [Given]

∴ T is the mid-point of GN [by theorem-2]

Again, in AGLN, T is the mid-point of GN and TM || GL (Given)

∴ M, is the mid-point of LN [by theorem-2]

∴ LM = MN. (Proved)

## Chapter 2 Theorems On Transversal And Mid Points Select The Correct Answer (MCQ)

**Question 1.**** In ΔPQR, ∠****PQR = 90° and PR = 10 cm. If S is the mid-point of PR, the length of QS is-**

- 4 cm
- 5 cm
- 6 cm
- 3 cm

**Solution:**

**Given**

**ΔPQR, ∠****PQR = 90° And PR = 10 cm**

**S is the mid-point of PR.**

Let us draw, SR’⊥ QR.

PQR = 90° and SR’R = 90°,

∴ PQ || SR.

Again, S is the mid-point of PR and SR’ || PQ.

∴ R’ is the mid-point of QR,

i.e., QR’ = RR’

Now, in A QSR’ and A RSR’, QR’ = RR’, SR’ is common to both the triangles and ∠SRQ = ∠SRR

[∵ each is 90° ]

ΔQSR’ = ΔRSR’

∴ QS = SR [∵ similar sides]

Now, PR 10 cm,

∴ SR = \(\frac{PR}{2}\) cm 5 cm.

∴ the required length of QS = 5 cm.

**Alternative Method:**

We know that the median of any triangle drawn from the vertex forming the right-angle of any right-angled triangle to its hypotenuse is half of the hypotenuse.

QS = \(\frac{1}{2}\) PR = \(\frac{1}{2}\)x10 cm = 5 cm.

∴ the required length of QS = 5 cm

**Question 2. ****The mid-point of the median AD of AABC is E. Extended BE intersects AC at F. If AC = 10-5 cm, then AF =**

- 3 cm
- 3.5 cm
- 2.5 cm
- 5 cm

**Solution:**

**Given**

**The mid-point of the median AD of AABC is E.**

**Extended BE intersects AC at F. If AC = 10-5 cm.**

Let us draw EF || DP.

Also, let DP intersects AC at P.

E is the mid-point of AD and DP || BF.

∴ P is the mid-point of CF, i.e., CP – \(\frac{1}{2}\) CF = PF

Now, AF + FP+ PC = AC

or, AF+AF+ PF AC [ FP= AF and PC = PF]

or, AF+AF+AF= AC.

or, 3 AF = AC

or, AF = \(\frac{1}{3}\)AC = \(\frac{1}{3}\) x 10.5 cm = 3.5 cm

∴ the required length, AF = 3.5 cm.

**Question 3.** **In the trapezium ABCD, AB || DC and AB =7 cm and DC=5 cm. If E and F be the mid-points of AD and BC respectively, then EF =**

- 5 cm
- 6 cm
- 7 cm
- 12 cm

**Solution:**

**Given**

**In the trapezium ABCD, AB || DC and AB =7 cm and DC=5 cm.**

**E and F be the mid-points of AD and BC respectively.**

Let us join B and D. Let BD intersects EF at P.

∴ DC || EF || AB and E and F are the mid-points of AD and BC respectively.

∴ EP = \(\frac{1}{3}\) AB and FP = \(\frac{1}{2}\) DC

∴ EF = EP + PF = \(\frac{1}{2}\) AB+ \(\frac{1}{2}\) DC = \(\frac{1}{2}\) (AB+DC)

= \(\frac{1}{2}\) (7+5) cm = \(\frac{1}{2}\) x 12 cm

= 6cm

∴ the required length of EF = 6 cm.

**Question 4.** **D, E, and F are the mid-points of BC, CA, and AB respectively of the AABC. If BE and DF intersect at X and CF and DE intersect at Y, then XY **

- \(\frac{1}{2}\) BC
- \(\frac{1}{4}\) BC
- \(\frac{1}{3}\) Bc
- \(\frac{1}{8}\) BC

**Solution:**

**Given**

**D, E, and F are the mid-points of BC, CA, and AB respectively of the AABC.**

**BE and DF intersect at X and CF and DE intersect at Y.**

D and E are the mid-points of BC and AC ofΔABC respectively.

∴ DE || BA and DE = \(\frac{1}{2}\) BA = BF………..(1)

Similarly, DF || CA and DF = CA = \(\frac{1}{2}\) CA = CE……………(2)

Now, in ΔBXF and ΔDXE,

∠XBF = ∠XED [ ∵ alternate angles]

<XFB = ∠XDE [∵ alternate angles ] and BF = DE [by (1)]

∴ ΔBXF ≅ΔDXE.

∴ BX=XE [ ∵ similar sides of congruent Δ’s]

∴ X is the mid-point of BE.

Similarly, it can be proved that Y is the mid-point of DE.

∴ in ABDE, X and Y are the mid-points of BE and DE respectively.

∴ XY = \(\frac{1}{2}\) BD = \(\frac{1}{2}\) x \(\frac{1}{2}\) BC

[∵ BD = \(\frac{1}{2}\) BC] = \(\frac{1}{4}\)BC.

∴ the required length of XY = \(\frac{1}{4}\)BC

**Question 5. ****D is the mid-point of the side BC of AABC. The line segment ****through D and parallel to CA and BA intersects BA and CA at E and F respectively. Then EF =**

- \(\frac{1}{4}\)BC
- \(\frac{1}{2}\) BC
- BC
- 2 BC

**Solution:**

**Given **

**D is the mid-point of the side BC of AABC.**

**The line segment ****through D and parallel to CA and BA intersects BA and CA at E and F respectively.**

DF || BA, F is the mid-point of AC and DF = \(\frac{1}{2}\)AB ……………..(1)

Again, DE | CA, E is the mid-point of AB, and DE = \(\frac{1}{2}\) AC.

∴ E is the mid-point AB,

∴ \(\frac{1}{2}\) AB= BE

or, DF = BE [by (1)]

∴ in the quadrilateral BDFE, BE || DF and BE = DF,

∴ BDFE is a parallelogram.

∴ EF = BD = \(\frac{1}{2}\) BC [ D is the mid-point of BC]

∴ the required length of EF = = BC.

**Question 6. E is the mid-point of the side BC of the parallelogram ABCD. DE and the extended AB meet at point F. Then AF =**

- \(\frac{3}{2}\)AB
- 2 AB
- 3 AB
- \(\frac{5}{4}\) AB

**Solution:**

**Given **

**E is the mid-point of the side BC of the parallelogram ABCD. DE and the extended AB meet at point F.**

AF || DC and BC are their transversals,

∠FBC = ∠DCB [alternate angles ]

or, <FBE = ∠DCE………..(1)

Again, AF | DC and DF are their transversal,

∴ ∠AFD = ∠CDF [alternate angles ]

or, <BFE = ∠CDE………….(2)

Now, in ABEF and ACDE,

<FBE = ∠DCE [ by (1)],

∠BFE = ∠CDE [ by (2)]

and BE = CE [E is the mid-point of BC]

ΔBEF ≅ ΔCDE. [by the condition of A-A-S congruence]

∴ BF = DC [∵similar sides of congruent triangles ]

∴ BF = AB………….(3) [ opposite sides of the parallelogram ABCD are equal, ∴ DC = AB. ]

Now, AF = AB + BF = AB + AB [by (3)] = 2AB

∴ AF = 2AB.

## Chapter 2 Theorems On Transversal And Mid Points Short Answer Type Questions

**Question 1.** **AD and BE are the medians of AABC and DF, parallel to BE, intersect AC at a point F. If the length of AC is 8 cm, find the length of CF.**

**Solution:**

**Given **

**AD and BE are the medians of AABC and DF, parallel to BE, intersect AC at a point F.**

**The length of AC is 8 cm.**

∴ D is the mid-point of BC and DF BE.

∴ F is the mid-point of CE.

Now, CF = \(\frac{1}{2}\) CE = \(\frac{1}{2}\) x \(\frac{1}{2}\) AC

[∵ BE is the median, E is the mid-point of AC]

= \(\frac{1}{4}\) AC = \(\frac{1}{4}\) x 8 cm

= 2 cm.

The length of CF = 2 cm.

**Question 2. P, Q, and R are the mid-points of the sides BC, CA, and AB of the AABC. If AC = 21 cm, BC = 29 cm, and AB = 30 cm, then find the perimeter of the quadrilateral ARPQ.**

**Solution:**

**Given **

**P, Q, and R are the mid-points of the sides BC, CA, and AB of the AABC.**

**AC = 21 cm, BC = 29 cm, and AB = 30 cm.**

AR = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 30 cm = 15 cm

RP = AQ [∵ ARPQ is a parallelogram, RP || AQ and PQ ||RA]

= \(\frac{1}{2}\) AC

= \(\frac{1}{2}\) x 21 cm

= 10.5 cm

∴ the perimeter of the quadrilateral

ARPQ=AR+RP + PQ+QA.

=AR+RP + AR+RP [ ∵ PQ-AR, QA=RP]

=2 (AR+ RP)=2(15+10-5) cm

=2 x 25.5 cm

= 51 cm

∴ the required perimeter = 51 cm.

**Question3. ****The medians BE and CF of the AABC intersect each other at G. P and Q are the mid-points of** **BG and CG respectively. If PQ = 3 cm, find the length of BC.**

**Solution:**

**Given**

**The medians BE and CF of the AABC intersect each other at G. P and Q are the mid-points of** **BG and CG respectively.**

**PQ = 3 cm**

In A BGC, P and Q are the mid-points of BG and CG respectively,

∴ PQ || BC and PQ = \(\frac{1}{2}\) BC.

Given that PQ-3 cm.

∴ 3 = \(\frac{1}{2}\) BC

or, BC= 3 x 2 cm

= 6 cm

**The length of BC = 6 cm**

**Question 4.** **D is any point on the side AC of the AABC. P, Q, X, Y are the mid-points of AB, BC, AD and DC respectively. If PX = 5 cm, then find the length of QY.**

**Solution:**

**Given **

**D is any point on the side AC of the AABC. P, Q, X, Y are the mid-points of AB, BC, AD and DC respectively.**

**PX = 5 cm.**

** **Let us join P and Q.

PQ || AC, [ ∵ P and Q are the mid-points of the sides AB and BC respectively]

or, PQ || XY and PQ = \(\frac{1}{2}\) AC ……..……..…(1)

Now, XY = DX+DY= \(\frac{1}{2}\) AD + \(\frac{1}{2}\) DC

= \(\frac{1}{2}\) (AD + DC)

= \(\frac{1}{2}\) AC…………………(2)

From (1) and (2) we get, PQ = XY

in the quadrilateral PXYQ, PQ = XY and PQ II XY

∴ PXYQ is a parallelogram.

PX=QY or, 5 cm=QY [∵ PX = 5 cm]

∴ QY = 5 cm.

The length of QY = 5 cm

**Question 5.** **D and E lie on the AABC such that AD = \(\frac{1}{4}\) AB and AE = \(\frac{1}{4}\) ****AC. Prove that DE || BC and ****DE = \(\frac{1}{4}\) BC.**

**Solution:**

**Given**

**D and E lie on the AABC**

**AD = \(\frac{1}{4}\) AB and AE = \(\frac{1}{4}\) ****AC.**

Let P and Q be the mid-points of AB and AC respectively and let us join P and Q.

Then, PQ || BC and PQ = \(\frac{1}{2}\) BC ………………(1)

Again AD = \(\frac{1}{4}\) AB = \(\frac{1}{2}\) x \(\frac{1}{2}\) AB = \(\frac{1}{2}\) AP [∵ \(\frac{1}{2}\) AB = AP]

AE = \(\frac{1}{4}\) AC = \(\frac{1}{2}\) x \(\frac{1}{2}\) AC = \(\frac{1}{2}\) AQ [∵ \(\frac{1}{2}\) AC = AQ]

Then, in A APQ, D, and E are the mid-points of AP and AQ respectively,

DE || PQ and DE = \(\frac{1}{2}\) PQ

or, DE || BC and DE = \(\frac{1}{2}\) x \(\frac{1}{2}\) BC [by (1)] = \(\frac{1}{4}\)BC

∴ DE || BC and DE = \(\frac{1}{4}\) BC. (Proved)

**Question 6.** **D, E, and F are the mid-points of the sides BC, CA, and AB respectively of the AABC. FE intersects AD at O. If AD = 6 cm, find the length of AO.**

**Solution:**

**Given **

**D, E, and F are the mid-points of the sides BC, CA, and AB respectively of the AABC.**

**FE intersects AD at O.**

**AD = 6 cm.**

D and E are the mid-points of BC and CA respectively.

∴ DE | BA or, DE || FA……………(1)

Again, D and F are the mid-points of BC and BA respectively,

∴ DF || CA

or, DF || EA……..(2)

From (1) and (2) we get, AEDF is a parallelogram, and AD and FE are two diagonals.

We know that diagonals of a parallelogram bisect each other.

∴ AO = DO

Now, AO + OD = AD

or, AO + AO = AD [∵ OD=AO]

or, 2AO = 6 cm

or, AO = \(\frac{6}{2}\)cm = 3 cm.

∴ the required length of AO = 3 cm.

The length of AO = 3 cm.

## Chapter 2 Theorems On Transversal And Mid Points Long Answer Type Questions

**Question 1. X and Z are the mid-points of the sides QR and QP respectively of the APQR. The side QP is extended to S such that PS = ZP. SX intersects PR at Y. Prove that PY= \(\frac{1}{4}\) PR.**

**Given:** X and Z are the mid-points of the sides QR and QP respectively of the APQR.

The side QP is extended to S such that PS = ZP. SX intersects PR at Y.

**To prove : **PY = \(\frac{1}{4}\) PR

**Proof**: ∵ X and Z are the mid-points of QR and QP respectively of the ΔPQR,

∴ XZ || RP and XZ = \(\frac{1}{2}\) PR …………..(1)

Again, ∵ ZP = PS,

∴ P is the mid-point of ZS.

PY is part of PR,

∴ ZX || PR ⇒ PY || ZX

in ΔSZX, PY || ZX and P is the mid-point of ZS.

PY =\(\frac{1}{2}\) ZX = \(\frac{1}{2}\) x \(\frac{1}{2}\) PR [by (1)]

∴ PY = \(\frac{1}{4}\) PR. (Proved)

**Question 2.** **Prove that the quadrilateral formed by joining the mid-points of a parallelogram is also a parallelogram.**

**Given:** E, F, G, and H are the mid-points of the sides AB, BC, CD, and DA respectively of the parallelogram ABCD.

A quadrilateral EFGH is constructed by joining the pair of points (E, F), (F, G), (G, H), and (H. E).

**To prove:** EFGH is a parallelogram.

**Construction:** Let us draw the diagonal AC of the parallelogram ABCD.

**Proof:** In Δ ABC, E, and F are the mid-points of BA and BC respectively,

∴ EF || AC and EF = \(\frac{1}{2}\) AC…………….(1)

Again, in Δ ACD, H, and G are the mid-points of DA and DC respectively,

∴ HG || AC and HG = \(\frac{1}{2}\) AC……………(2)

Now, from (1) and (2) we get, EF = HG and EF || HG, i.e., two opposite sides of the quadrilateral EFGH are equal and parallel.

Hence, EFGH is a parallelogram. (Proved)

**Question 3. Prove that the quadrilateral formed by joining the mid-points of the sides of a square is also a square.**

**Given:** Let P, Q, R, and S be the mid-points of the sides AB, BC, CD, and DA respectively of the square ABCD.

A quadrilateral PQRS is constructed by joining these points.

**To prove:** PQRS is a square.

**Construction: **Let us draw the diagonals AC and BD of the square ABCD and let AC and BD intersect each other at O and also let AC intersect PS at E, and BD intersect PQ at F.

**Proof:** In ΔABD, S and P are the mid-points of AD and AB respectively,

∴ SP || DB and SP = \(\frac{1}{2}\) DB ……………(1)

Again, ΔBCD, Q, and R are the mid-points of CB and CD respectively,

∴ RQ | DB and RQ = \(\frac{1}{2}\) DB…………….(2)

Now, from (1) and (2) we get, SP = RQ and SP || RQ.

∴ PQRS is a parallelogram.

Again, in A ABC, P, and Q are the mid-points of AB and BC, PQ || AC and PQ=AC….(3)

SP = \(\frac{1}{2}\) DB and PQ = \(\frac{1}{2}\) AC.

But, DB = AC,

∴ SP = PQ, i.e.,

in quadrilateral EPFO, EP || OF and PF || EO.

∴ EPFO is a parallelogram.

∴ ∠EPF = ∠EOF [ ∵ opposite angles of a parallelogram]

= 90° [∵ the diagonals of a square bisect each other orthogonally. ]

∴ PQRS is such a rhombus, one angle of which is a right angle,

∴ PQRS is a square. (Proved)

**Question 4.** **Prove that the quadrilateral formed by joining the mid-points of a rhombus is a rectangle. **

**Given: **P. Q, R, and S are the mid-points of AB, BC, CD, and DA respectively of the rhombus ABCD.

A quadrilateral PQRS is constructed by joining these points.

**To prove:** PQRS is a rectangle.

**Construction:** Let us draw the diagonals AC and BD of the rhombus ABCD. Also, let AC and BD intersect each other at O and AC intersect SP at E, and BD intersects PQ at F.

**Proof:** In A ABD, S, and P are the mid-points of AD and AB respectively, SP || DB and SP = \(\frac{1}{2}\) DB……….(1)

Again, in ΔBCD, R and Q. are the mid-points of CD and CB respectively, RQ || DB and RQ=DB …………..(2)

From (1) and (2) we get, SP || PQ and SP = RQ.

∴ PQRS is a parallelogram.

Now, in ΔABC, P, and Q are the mid-points of AB and BC respectively,

PQ || AC and PQ = \(\frac{1}{2}\) AC……………..(3)

From (1) and (3) we get, SP = \(\frac{1}{2}\) DB and PQ = \(\frac{1}{2}\)AC

But DB ≠ AC

∴ SP PQ, i.e., two adjacent sides of the parallelogram PQRS are not equal.

∴ PQRS is neither a rhombus nor a square.

Again, in the quadrilateral PEOF, PE || FO and EO || PF.

∴ PEOF is a parallelogram.

:. ∠EPF = ∠EOF [ ∵ opposite angles of a parallelogram]

= 90° [ ∵ the diagonals of a rhombus bisect each other orthogonally. ]

∴ PQRS is a parallelogram, one of the angles of which is a right angle, but the sides of which are not equal.

∴ PQRS is a rectangle. (Proved)

**Question 5.** **D and E are the mid-points of AB and AC respectively of the ΔABC. P and Q are the mid-points of CD and BD respectively. Prove that BE and PQ bisect each other.**

**Given:** D and E are the mid-points of AB and AC respectively of the Δ ABC.

P and Q are the mid-points of CD and BD respectively. Let us join B, E, and P, Q.

**To prove:** BE and PQ bisects each other.

**Construction: **Let us join P, E; Q, E, and B, P.

**Proof:** In ΔACD, P, and E are the mid-points of CD and CA respectively,

PE || DA and PE = \(\frac{1}{2}\) DA……….(1)

Again, DA, BA, and BQ are the line segments of the same straight line

∴ PE || BQ.

Now, D is the mid-point of AB.

∴ DA = DB or, DA = DQ + QB

or, DA=BQ+ BQ [∵ Q is the mid-point of BD]

or, DA = 2 BQ

∴from (1) we get, PE = \(\frac{1}{2}\) DA =\(\frac{1}{2}\) × 2BQ = BQ.

in the quadrilateral PEQB, PE || BQ and PE = BQ, i.e., two of the opposite sides of PEQB are equal and parallel.

PEQB is a parallelogram, and BE and PQ is two of its diagonals.

∴ BE and PQ bisects each other. (Proved)

**Question 6. AD is a perpendicular drawn from A to the bisector of the ∠ABC. The line segment DE is drawn through D, parallel to the line segment BC, which intersects AC at E. Prove that AE = EC.**

**Given:** AD is the perpendicular, drawn from A to the bisector of ∠ABC.

The line segment DE is drawn through D, parallel to the line segment BC, which intersects AC at E.

**To prove** AE = EC.

**Construction:** Let us extend AD to a point F which intersects BC at F.

**Proof:** In Δ ABD and ΔFBD. ∠ABD = ∠FBD [∵ BD is the bisector of ∠ABC]

∠ADB = ∠FDB [ each is right-angle] and BD is common to both the triangles.

∴ ΔABD ≅ ΔFBD [ by the condition of A-A-S congruence]

AD = DF [∵ similar sides of congruence triangles ]

∴ D is the mid-point of AF.

Again, DE || BC (Given)

or, DE || FC. [∵ BC and FC are the segments of the same straight line]

Then, in ΔAFC, D is the mid-point of AF and DE || FC.

∴ E is the mid-point of AC.

∴ AE = EC. (Proved)

**Question 7.**** AD is a median of AABC. Line segments BR and CT through B and C respectively are drawn parallel to AD, which intersects extended BA at T and extended CA at R. Prove \(\frac{1}{\mathrm{AD}}=\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}\) that**

**Given:** AD is a median of AABC. The line segments BR and CT through B and C respectively are drawn parallel to AD which intersects extended BA at T and extended CA at R.

**To prove : **\(\frac{1}{\mathrm{AD}}=\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}\)

AD RB TC

**Proof:** In ABCR, D is the mid-point of CB and DA || BR (by construction)

∴ A is the mid-point of CR and DA = \(\frac{1}{2}\) BR or, RB = 2 AD………(1)

Again, in ΔBCT, D is the mid-point of BC and DA||CT (by construction)

A is the mid-point of BT and DA = \(\frac{1}{2}\) CT

or, TC = 2 AD……….(2)

\(\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}=\frac{1}{2 \mathrm{AD}}+\frac{1}{2 \mathrm{AD}}\) [from (1) and (2)]

= \(\frac{1+1}{2 \mathrm{AD}}=\frac{2}{2 \mathrm{AD}}=\frac{1}{\mathrm{AD}}\)

∴ \(\frac{1}{\mathrm{AD}}=\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}\) (Proved)

**Alternative Method:**

\(\frac{1}{\mathrm{AD}}=\frac{2}{2 \mathrm{AD}}=\frac{1+1}{2 \mathrm{AD}}=\frac{1}{2 \mathrm{AD}}+\frac{1}{2 \mathrm{AD}}=\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}\) [by (1) and (2) ]

∴ \(\frac{1}{\mathrm{AD}}=\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}\) (proved)

**Question 8.** **In the trapezium ABCD, AB || DC and AB > DC; E and F are the mid-points of the ****diagonals AC and BD respectively. Prove that EF = \(\frac{1}{2}\) (AB-DC).**

**Given:** In the trapezium ABCD, AB || DC and AB > DC; E and F are the mid-points of the diagonals AC and BD respectively.

**To prove: **EF = \(\frac{1}{2}\) (AB-DC).

**Construction:** The line segment DE is extended to G, which intersects AB at G.

**Proof:** In ΔAEG and ΔCED, ∠GAE = alternate ∠DCE [∵ AB || DC and AC are their transversal ]

∠AGE = alternate ∠CDE [ ∵ AB || DC and DG is their transversal ].

and AE = CE [∵ E is the mid-point of the diagonal AC]

:. ΔAEG = ΔCED

∴ AG=CD…………(1) [∵ similar sides of congruent triangles] and DE GE, i.e., E is the mid-point of DG.

Now, in ΔBDG, F and E are the mid-points of BD and GD respectively.

∴ FE BG and FE = \(\frac{1}{2}\) BG.

∴ EF = BG = \(\frac{1}{2}\) (AB-AG) = \(\frac{1}{2}\) (AB-CD) [by (1) AG = CD]

∴ EF = \(\frac{1}{2}\) (AB-DC). (Proved)

**Question 9.** **C is the mid-point of the line segment AB and PQ is any straight line. The least distances of PQ from A, B and C are AR, BS and CT respectively. Prove that AR + BS = 2 CT.**

**Given:** C is the mid-point of the line segment AB and PQ is any straight line.

The least distances of PQ from A, B, and C are AR, BS, and CT respectively.

**To prove **AR + BS = 2 CT.

**Construction: **Let us join A and ‘S and let AS intersects CT at D.

**Proof:** Since the least distances of PQ from A, B, and C are AR, BS, and CT respectively, AR, BS, and CT are all perpendiculars to PQ.

∴ AR || BS || CT.

Now, in ΔABS, CD || BS,

∴ CD = \(\frac{1}{2}\) BS [∵ C is the mid-point of AB and D is the mid-point of AS]

or, BS = 2CD…………(1)

Again, in ΔARS, DT || AR,

∴ DT = \(\frac{1}{2}\) AR [ ∵ D is the mid-point of AS ]

or, AR = 2 DT ………..(2)

Now, AR + BS2 DT+2 CD [∵ from (1) and (2)]

= 2 (DT + CD)

= 2 CT

∴ AR + BS = 2CT. (Proved)

**Question 10. ABCD is a square. The diagonals AC and BD of it intersects each other at O. The bisector of <BAC intersects BO at P and BC at Q. Prove that OP = \(\frac{1}{2}\) CQ.**

**Given:** The diagonals AC and BD of the square ABCD intersect each other at O.

The bisector of <BAC intersects BO at P and BC at Q.

**prove : **OP = \(\frac{1}{2}\) CQ.

**Construction: **Let us draw DB || CR. CR intersects the extended AQ at R.

Proof: O is the mid-point of AC [∵ diagonals of a rhombus intersect each other orthogonally.]

and OP || CR [by construction]

∴ OP = CR………….(1)

Now, ∠BAQ = <CAR (Given),

∠ABQ = ∠AOP [ each are 90° ]

= similar ACR [ ∵ OP || CR and AC is their transversal ]

∴ in ΔABQ and ΔACR, ∠BAQ= ∠CAR and ∠ABQ = ∠ACR.

∴ ∠AQB = ∠ARC [ third angles ], But ∠AQB = ∠CQR [opposite angles]

∴ ∠ARC=∠CQR

or, ∠QRC = ∠CQR.

∴ CR = CQ. [∵ opposite sides of equal angles. ]

∴ From (1) we get, OP = \(\frac{1}{2}\) CQ [∵ CR=CQ] (Proved)