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Home / Geometry / Prove that, summation of four angles of a quadrilateral is equal to four right angles.

# 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.

AF=CF          [F is midpoint of AC]

DF=OF        [by construction]

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

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

And ∠DAF=∠FCO

Or, ∠DAC=∠ACO

But these are alternate angles

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]

AB=CD         [given]

AC=AC      [common side]

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

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.

(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.

AB∥CD   [Given]

∴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=6a2 sq. unit

=6(6.5)2 sq.cm

=6×42.25 sq.cm

=253.5 sq.cm (Ans.)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...