Question

state and prove basic proportionality theorem

Answer 1

Basic Proportionality Theorem: A line parallel to one side of a triangle divides the other two sides into parts of equal proportion.

Given that In triangle ABC, a line drawn parallel to BC cuts AB and AC at P and Q respectively. See the attachment.
To Prove: AP/PB = AQ/QC

Proof:
In ΔABC and ΔAPQ
Since PQ is parallel to BC
∠ABC = ∠APQ
∠ACB = ∠AQP
∠BAC = ∠PAQ

Hence  ΔABC ~ ΔPAQ
Since they are similar, the ratio of their sides are equal.

$\frac{AB}{AP} = \frac{AC}{AQ}\\ \\ \Rightarrow \frac{AB}{AP}-1 = \frac{AC}{AQ}-1\\ \\ \Rightarrow \frac{AB-AP}{AP} = \frac{AC-AQ}{AQ}\\ \\ \Rightarrow \frac{PB}{AP} = \frac{QC}{AQ}\\ \\ \Rightarrow \frac{AP}{PB} = \frac{AQ}{AC}$

Hence, Proved!

Answer 2

hi mate,

PROOF OF BPT

Given: In ΔABC, DE is parallel to BC

Line DE intersects sides AB and AC in points D and E respectively.

To Prove:

AD AE

----- = -----

DB AC

Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.

Proof:

Area of Triangle= ½ × base × height

In ΔADE and ΔBDE,

Ar(ADE) ½ ×AD×EF AD

----------- = ------------------ = ------ .....(1)

Ar(DBE) ½ ×DB×EF DB

In ΔADE and ΔCDE,

Ar(ADE) ½×AE×DG AE

------------ = --------------- = ------ ........(2)

Ar(ECD) ½×EC×DG EC

Note that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.

So, we can say that

Ar(ΔDBE)=Ar(ΔECD)

Therefore,

A(ΔADE) A(ΔADE)

------------- = ---------------

A(ΔBDE) A(ΔCDE)

Therefore,

AD AE

----- = -----

DB AC

Hence Proved.