**Telanagana SCERT Solution Class IX (9) Math Chapter 8 Quadrilaterals Exercise 8.4**

** (1) ABC is a triangle. D is a point on AB such that AD = 1/4AB and E is a point on AC such that AE = 1/4AC. If DE = 2 cm find BC**

**Solution:** Given,

ABC is a triangle

AD = 1/4AB, AE = 1/4AC

DE = 2 cm

Let, Us, consider point F and G mid points of side AB & AC respectively

In △ABC

FG = ½ BC [∵ F and G are midpoints of AB & AC and FG equal to ½ BC which is the third side of the triangle]

Now,

In △AFG

D and E are the midpoints of AF & AG respectively

Since, AD is ¼ of AB & AE is ¼ of AC

And F & G being the mid points of AB & AC respectively

DE = ½ FG [D & E are mid points of AF and AG]

Or, DE = ½ x ½ BC,

Or, DE = ¼ BC (proved)

**(2) ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.**

**Solution:** Given,

ABCD is a quadrilateral

E, F, G, H are the mid points of AB, BC, CD and DA respectively

Join,

AC & BD

∴ In △ADC

H & G are mid points of AD & DC respectively

∴ By mid point theorem

HG ∥ AC & HG = ½ AC

Similarly In △ABC

E & F are mid points of AB & BC respectively

∴ By mid point theorem

EF ∥ AC & EF = ½ AC

∴ HG ∥ EF & HG = EF [∵ both are parallel to AC & both are ½ of AC]

Now,

In △ABD

H & E mid points of AD & AB respectively

∴ By Mid points theorem

EH ∥ BD & EH = ½ BD

Similarly, In △BCD

By mid point theorem

FG ∥ BD & FG = ½ BD

∴ EH ∥ FG & EH = FG [∵ both are parallel to BD & both are ½ of BD]

∴ Quad EFGH is a parallelogram since opposite sides are parallel to each other and equal

**(3) Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.**

**Solution:** Given,

ABCD is a rhombus

E, F, G, H are mid points of side AB, DC, CD, DA respectively

In △ABD

E, F are mid points of side AB and AD respectively

By mid points theorem

EF ∥ BD & BF = ½ BD

In, △ABC

E, F are mid points of side AB & BC respectively

By mid points theorem,

EH ∥ AC & EH ½ of AC

We know,

Diagonals of rhombus are perpendicular to each other,

∴ ∠POS = ∠POQ = ∠QOR = ∠SOR = 90^{o}

∠AOB = 90^{o}

∠APE = ∠PES = 90^{o} [alternate angles EH AC]

In △ADC

F & G are mid points of side AD & DC respectively

By mid points theorem

FG ∥ AC & FG = ½ AC

Now, ∠APF = ∠AOD = 90^{o}

∠ADF = ∠PFQ = 90^{o}

[Alternate angles, FG ∥ AC]

[∵ Corresponding angles BF BD & BD & AC are diagonals of rhombus]

Now,

In △BCD

H, G are the mid points of sides BC and DC respectively,

∴ By mid points theorem,

HG ∥ BD & HG = ½ BC

∴ Now,

∠QOR = ∠CRG = 90^{o} [∵corresponding angle HG ∥ BD]

∠QGR = ∠CRG = 90^{o} [alternate angles, FG ∥ AC]

Similarly, ∠SHR = ∠CRH = 90^{o} [alternate angle, EF ∥ AC]

∴ In quad EFGH

∠SHR = ∠QGR = ∠PFQ = PES = 90^{o}

And

Opposite sides EF & GH ∥ BD and equal to ½ BD

Opposite sides FG & EH ∥ AC & equal to ½ AC

∴ Opposite sides of equal EFGH are equal and parallel to each other

And each vertice angles of quad EFGH is are 90^{o}

∴ Quad EFGH is rectangle (proved)

**(4) In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.**

**Solution:** Given,

ABCD is a parallelogram

E and F are mid points of AB & DC respectively

∴ AE = ½ AB

FC = ½ DC

But,

AB = DC [∵ opposite sides of parallelogram ABCD if equal and parallel to each other]

AB ∥ DC

∴ AE = FC [∵ parts of parallel sides are parallel to each other]

& AE ∥ FC

∴ Quad AEFC is a parallelogram

∴ AF ∥ EC

Now,

In △APB

QE ∥ AP [∵ parts of parallel lines are parallel to each other, AF ∥ EC]

E is the mid point of side AB

∴ By mid point theorem,

Q is the mid point of PB since QE ∥ AP

∴ PQ = BQ

Now,

In △DQC

PE ∥ CQ [∵ parts of parallel lines are parallel to each other, EC ∥ AF]

F is the mid point of side DC

∴ By mid point theorem,

P is the mid point of DQ since PE ∥ CQ

∴ DP = PQ

∴ DP = PQ = BQ

Hence AF and CE trisects diagonal BD

**(5) Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other**

ABCD is a quadrilateral

E, F, G, H are the mid points of AD, DC, CB, BA respectively

In △ABD

E and H are mid points of AD & AB respectively

∴ By mid point theorem,

EH ∥ BD & EH = ½ BD – (i)

In △BCD

G and F are mid points of BC & Dc respectively

∴ By mid points theorem,

GF ∥ BD & GF = ½ BD – (ii)

∴ On Comparing equation (i) & (ii) we get,

EH ∥ GF & FH = GF [∵ both equal to ½ BD and parallel to BD]

∴ Quad EFGH is a parallelogram

Hence, EG and FH are diagonals of parallelogram EFGH

∴EG and FH bisects each other

Line segments EG and FH joining mid points of sides AD, BC and AB, DC of quad ABCD bisects each other

**(6) ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that **

**(i) D is the midpoint of AC **

**(ii) MD ****⊥**** AC (iii) CM = MA = 1/2AB**

**Solution: **Given,

ABC is a triangle right angled at C

A line through mid point of AB which is M lies parallel to BC and meet AC at D

In △ABC

(i) M is the mid point of AB

DM ∥ BC [given]

∴ By mid point theorem,

D is the mid point of AC [∵ mid point when connected are parallel to the third side]

∠MCE = ∠DMC [alternate angles, DM BC]

∠DCM = ∠ACB – ∠MCB [∠ACB = 90^{o}]

Or, ∠DCM = 90 – ∠ACB

Or, ∠DCM = 90 – ∠DMC [∠MCB = ∠DMC] – (i)

In △DCM

∠MDC + ∠DMC + ∠DCM =180^{o} [sum of angles of a triangle is 180^{o}]

Or, ∠MDC + ∠DMC + 90^{o} – ∠DMC = 180^{o} from equation (i)

Or, ∠MDC = 180^{o} – 90^{o}

= 90^{o}

(ii) ∴ Line segment MD is perpendicular to AC

Since angles between the two segments is 90^{o}

(iii) In △ADM & △CDM

(i) DM is common side

(ii) ∠MDC + ∠ADM = 180^{o} [collinear angles]

Or, 90^{o} + ∠ADN = 180^{o}

Or, ∠ADM = 180^{o} – 90^{o}

Or, ∠ADM = 90^{o}

∠ADM = ∠MDC = 90^{o}

(iii) AD = DC [∵ D is the mid point of AC]

∴ △ADM ≅ △CDM by SAS

∴ CM = MA [corresponding sides of congruent triangles are equal]

And

MA = ½ AB [M is the third point of AB]

∴ CM = NA = ½ AB proved