# Basic Proportionality Theorem

**Basic proportionality theorem:**

**Converse of basic proportionality theorem:**

In a triangle ABC, D and E are points on the sides AB and AC respectively such that DE || BC. IF AD = 3cm, BD = 2cm, AE = 2.7cm find AC.

Let EC =

*x*. AC =?

We know,

3* Ã—*

*x*= 2.7 Ã— 2

*x*=

= 0.9 Ã— 2

= 1.8 cm

AC = AE + EC

= 2.7 + 1.8 cm

= 4.5 cm

The diagonals of a quadrilateral ABCD cut at K. If AK = 2.4 cm KC = 1.6 cm, BK = 1.5 cm, KD = 1 cm. Prove that AB || DC.

AK = 2.4 cm, BK = 1.5 cm, CK = 1.6cm, and DK = 1 cm

â‡’

â‡’ AB || CD.

If OA = 12 cm

AB = 9 cm

OC = 8 cm

EF = 4.5 cm

Find CD and OF, if AC || BD

and CE || DF

Let CD =*x*

*OE =*

*y*

Since AC || BD,

12

*Ã—*

*x*= 9 Ã— 8

*x*=

= 3 Ã— 2

= 6 cm

âˆ´ CD = 6 cm

Since CE || DF,

6 Ã—

*x*= 8 Ã— 4.5

*y*=

= 4 Ã— 1.5

OE = 6 cm

OF = OE + EF = 6 + 4.5 = 10.5 cm

Ans: CD = 6 cm

OF = 10.5 cm

In the given figure, if DE || BC and CD || EF. Prove that AD^{2} = ABÃ— AF

DE || BC, CD || EF To prove: Proof: CD || EF From (1) and (2), From (3) and (4) we get, AD Ã— AD = Ã— AD |

Prove that the line segment joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram.

Given:

ABCD is a quadrilateral. E, F, G and H are the mid-points of AB, BC, CD and DA respectively.

To prove:

EFGH is a parallelogram

Construction:

Join EF, FG, GH and HE. Join AC and DB.

Proof:

In the triangle ABC,

AE = EB ( Sine E is the mid-point of AB).

BF = FC (Since F is the mid-point of BC).

By basic proportionality theorem, AC is parallel to EF.

In the triangle ADC,

AH = HD (Since H is the mid-point of AD).

DG = GC (Since G is the mid-point of DC).

By basic proportionality theorem, GH is parallel to AC.

Since AC is parallel to EF, GH is parallel to EF.

Similarly, we can prove EH is parallel to GF.

âˆ´ EFGH is a parallelogram.

Prove that the diagonal of a trapezium divide each other proportionately.

Hint: From the point of intersection of the diagonals, draw the line parallel to the parallel sides to meet one of the non-parallel sides at a point.

Given: ABCD is a trapezium

__ To prove:__ The diagonals AC and BD divide each other proportionally.

Let O be the point of intersection of the diagonals.

Construction:

Through O draw a line OX || AB meeting BC at X.

Proof: In Î” ABC, OX || AB.

â‡’ ___ (1)

In Î” BCD, OX || CD

___ (2)

From (1) and (2)

**Example : **The diagonals of a trapezium divide each other proportionately.

Example 7: The side BC of a triangle ABC is bisected at D. O is any point in AD. BO and CO produced meet AC and AB in E and F respectively and AD is produced to G so that D is the mid point of OG.

Prove that AO: AG = AF: AB and that FE || BC.

Proof:

**Example 8: **Two circles touch each other internally at P. If ACP and BDP are the lines meeting the two circles at A, B and C, D respectively. Prove that

**Solution:**

__ Given:__ Two circles touch each other at P.

To prove:

**Proof:**

Using tangent chord theorem,

We get,

Also,

â‡’ (Since corresponding angles are equal)

â‡’ CD || AB

Using basic proportionality theorem in the triangle PAB,

Hence proved.