# Ex: In the figure, ABC is an equilateral triangle.

**D, E and the midpoint of AB,BC and AC respectively.**

**(a) Prove that, summation of four angles of a**

** quadrilaterals equal to four right angles.**

**Prove that, ∠BDF+∠DFE+∠FEB+∠EBD= 4 right angles. **

** (b) Prove that, DF∥BC and DF = ½ BC.**

**Particular enunciation**: Given that, in the figure, ABC is an equilateral triangle. D,E and F are the midpoints of AB, BC and AC respectively. We join D, F; E, F and D,E.

We have to prove that ∠BDF+∠DFE+∠FEB+∠EBD = 4 right angle.

**Proof**: In ∆DBE

∠DBE+∠BED+BDE = 2 right angle. ————— (1) [∵sum of three angles of a triangle is equal to 2 right angle]

Again, In ∆DEF

∠EDF+∠DEF+∠DFE= 2 right angle ———————- (2)

Adding (1) and (2) we get,

DBE+∠BED+∠BDE+∠EDF+∠DEF+∠DFE=(2+2)right angle

Or, ∠DBE+(∠BED+∠DEF)+ ∠EFD+(∠BDE+∠EDF)=4 right angle

Or, ∠DBE+∠BEF+∠EFD+∠BDF=4 right angle.

Therefore ∠BDF+∠DFE+∠FEB+∠EBD=4 right angle (Proved)

**Ex: Prove that, DF∥BC and DF= ½ BC**

**Particular enunciation**: given that in the figure, ABC is an equilateral triangle. D, E and F are the mid points of AB, BC and AC respectively. We join D and F.

We have to prove that DE∥BC and DF = ½ BC.

**Construction**: we produce DF upto O such that DF=FO. We join O and C.

**Proof**: In ∆ADF and ∆COF

AF=CF [F is midpoint of AC]

DF=OF [by construction]

And included ∠AFD=included ∠CFO [vertically opposite angles are equal to each other]

So ∆ADF≌∆COF [SSS theorem]

∴AD=CO

Or, BD=CO [[D is an midpoint AB]

And ∠DAF=∠FCO

Or, ∠DAC=∠ACO

But these are alternate angles

∴AD∥CO

Or, BD∥CO.

In ⎕BCOD,

The opposite sides BD and CO are equal and parallel.

∴⎕BCOD is a parallelogram.

So DO ∥BC [the opposite of a parallelogram are equal and parallel]

Or, DF ∥BC

And DO=BC

Or, DF+FO=BC

Or, DF+DF=BC [DF=FO]

Or, 2 DF=BC

∴DF= ½ BC (Proved)

**Ex: In the figure, AB=CD and AB∥CD.**

**(a)Name the two triangles on base AB.**

**(b)Prove that AD and BC are equal and**

**Parallel to each other**

**(c)Show that, AO=OC and OB=OD.**

Solution:** (a)**

From the figure, the two triangles on base AB are(i) ∆ABC and (ii)∆ABD.

**(b)**

**Particular enunciation** : Given that AC=CD. We have to prove that AD and BC are equal and parallel to each other.

Construction: we join A and C.

**Proof**: AB∥CD and AC is their Transversal.

∴∠BAC=alternative ∠DCA ————- (i) [Alternative angles are equal to each other]

In ∆ABC and ∆ADC

AB=CD [given]

AC=AC [common side]

And included ∠BAC=included ∠DCA [from (i)]

So, ∆ABC≌∆ ADC [SSS theorem]

∴ AD=BC

And∠ BCA=∠DAC

Now, AC is the transversal of the sides AD and BC and∠ BCA=alternate∠ DAC

We know, if alternate angles are equal then two sides are parallel.

∴AD∥BC (proved)

**(c)** **Particular enunciation**: given that AB=CD and AB∥CD. We join A,C and B,D.AC and BD intersects each other at O.

We have to prove that OA=OC and OB=OD.

Proof: In the quadrilateral ABCD

AB∥CD [Given]

And AD∥BC [from (b)]

∴ABCD is a parallelogram.

In the parallelogram ABCD, the diagonals AC and BD intersect each other at O.

Since, the diagonals of a parallelogram bisect each other

∴AO=OC and OB = OD (Proved)

**Ex: Length, breadth and height of a rectangular solid are 10cm, 8cm and 5cm respectively. Find the area of the whole face of the solid.**

Length of a rectangular solid a, =10cm

Breadth of a rectangular solid is b,=8cm

Height of a rectangular solid is c,=5cm

We know,

Area of the whole face of solid =2 (ab + bc + ca)sq. unit

=2(10×8+8×5+5×10)sq.cm

=2(80+40+50)sq.cm

=2×270 sq.cm

=340 sq.cm(Ans.)

**Ex: If the edge of a cubic box is 6.5 cm, find the area of the entire faces of the box.**

The edge of a cubic box, a=6.5 cm

Area of the entire faces of the box=6a^{2} sq. unit

=6(6.5)^{2} sq.cm

=6×42.25 sq.cm

=253.5 sq.cm (Ans.)