In ∆ABC , if DE || BC and intersects AB in D and AC in E then
AD AE ---- = ------ DB EC |

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

Statements |
Reasons |

1) EF ┴ BA | 1) Construction |

2) EF is the height of ∆ADE and ∆DBE | 2) Definition of perpendicular |

3)Area(∆ADE) = (AD .EF)/2 | 3)Area = (Base .height)/2 |

4)Area(∆DBE) =(DB.EF)/2 | 4) Area = (Base .height)/2 |

5)(Area(∆ADE))/(Area(∆DBE)) = AD/DB | 5) Divide (4) by (5) |

6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC | 6) Same as above |

7) ∆DBE ~∆DEC | 7) Both the ∆s are on the same base and between the same || lines. |

8) Area(∆DBE)=area(∆DEC) | 8) If the two triangles are similar their areas are equal |

9) AD/DB =AE/EC | 9) From (5) and (6) and (7) |

1) In the given figure, PQ || MN. If KP / PM = 4 /13 and KN = 20.4 cm.

Find KQ.

In Δ KMN,

PQ || MN

∴ KP / PM = KQ / QN ( By BPT theorem)

⇒ KP / PM = KQ / ( KN – KQ)

⇒ 4 / 13 = KQ / ( 20.4 – KQ)

⇒ 4( 20.4 - KQ ) = 13 KQ ( cross multiply )

⇒ 81.6 – 4KQ = 13 KQ

⇒ 17KQ = 81.6

⇒ KQ = 81.6 / 17

∴

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2) In the figure given below, DE ||BC. If AD = x cm, DB = x-2 cm,

AE = x-1 cm, then find the value of x.

In triangle ABC ,

DE || BC

AD /DB = AE /EC ( by basic proportionality theorem)

⇒ x /(x - 2) = (x + 2) /(x - 1)

⇒ x (x - 1) = (x - 2)(x + 2) (by cross multiplication)

⇒ x

⇒ -x = -4

∴ x = 4

• Similarity in Geometry

• Properties of similar triangles

• Basic Proportionality Theorem(Thales theorem)

• Converse of Basic Proportionality Theorem

• Interior Angle Bisector Theorem

• Exterior Angle Bisector Theorem

• Proofs on Basic Proportionality

• Criteria of Similarity of Triangles

• Geometric Mean of Similar Triangles

• Areas of Two Similar Triangles

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