__Basic Proportionality Theorem Class 10th __
__(OR) B.P.T. or Thales Theorem__

__ ____Converse of basic proportionality theorem, thales theorem 10th standard, theorem 6.2 class 10__

**Statement:- **

**If a line is drawn parallel to one
side of the triangle to intersect the other two sides in two distinct points, the other two sides are divided in the same ratio.**

**Given:- **

**A Δ ABC in which line ***l ***॥**** BC, intersect side AB and AC at point D and E**

**To Prove :- \[\frac{AD}{DB}=\frac{AE}{EC}\]**

**Construction :- **

**Draw EM ****丄**** AB and DN ****丄**** AC. Also join
BE and CD**

**Proof :- \[Area\: \: \: of\: \: triangle = \frac{1}{2}\times Base\times Height\]**

**\[Area\: \: \: of\: \:\Delta ADE=\frac{1}{2}\times AD\times EM............(1)\]**

**\[Area\: \: \: of\: \:\Delta BDE=\frac{1}{2}\times BD\times EM .......... (2)\]**

**\[Dividing\: \: equation \: \: (1)\: \: by\: \: equation\: \: (2)\]**

**\[\frac{Ar(\Delta ADE)}{Ar(\Delta BDE)}=\frac{\frac{1}{2}\times AD\times EM}{\frac{1}{2}\times BD\times EM}\]**

**\[\frac{Ar(\Delta ADE)}{Ar(\Delta BDE)}=\frac{AD}{BD}\: \: \: ....................(3)\]**

**Similarly**

**\[\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times EC\times DN}\]**

**\[\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}=\frac{AE}{CE}\: \: \: \: ....................(4)\]**

**ΔBED and ΔCDE have same base and lie
between the same parallel DE and BC**

**
****∴**** Ar(ΔBED) = Ar(ΔCDE)
...........................................(5)**

**From equation (3), (4), (5)
...........**

**\[\frac{Ar(\Delta ADE)}{Ar(\Delta BED)}=\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}\]**

**\[\frac{AD}{DB}=\frac{AE}{EC}\]**

**Hence prove the Basic Proportionality Theorem.**

**Note:- ***For the examination point of view students should study the basic proportionality theorem only its converse is only a motivational theorem. *

__Converse of Basic Proportionality Theorem__

**If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. **

**Given:- ****\[In\: \Delta ABC,\:\frac{AD}{DB}=\frac{AE}{EC}\]**

**
**
**To Prove :- ****Line ***DE*॥ BC

**Construction:-**

**If DE is not parallel to BC, then let us take another line DE'****॥**** BC**

**Proof:-**

**
**
**In ****ΔABC, ****DE'****॥**** BC ****Therefore by B.P.T**

**\[\frac{AD}{DB}=\frac{AE'}{E'C}\]**

**\[But\: \frac{AD}{DB}=\frac{AE}{EC}\: \: ............Given\]**

**\[Therefore\: \: \: \: \: \: \frac{AE}{EC}=\frac{AE'}{E'C}\]**

**\[1+\frac{AE}{EC}=1+\frac{AE'}{E'C}\]**

**\[\frac{EC+AE}{EC}=\frac{E'C+AE'}{E'C}\]**

**\[\frac{AC}{EC}=\frac{AC}{E'C}\]**

** This is possible only if E and E' coincide with each other**

**⇒ E and E' represent the same point on the side of the triangle.**

** Hence DE is parallel to the side BC**

**Converse of BPT is proved**

**Important result based on BPT**

**If a line intersects side AB and AC of a ****ΔABC at D and E respectively and is parallel to BC, then prove that ****\[\frac{AD}{DB}=\frac{AE}{AC}\]**

**Solution :- **

**It is given that DE॥BC, therefore by BPT****\[\frac{AD}{DB}=\frac{AE}{EC}\]**

**\[\frac{DB}{AD}=\frac{EC}{AE}\: \: \: .........By\: \: Invertendo\]**

**Adding 1 on both side we get**

**\[\frac{DB}{AD}+1=\frac{EC}{AE}+1\]**

**\[\frac{DB+AD}{AD}=\frac{EC+AE}{AE}\]**

**\[\frac{AB}{AD}=\frac{AC}{AE}\]**

**\[\frac{AD}{AB}=\frac{AE}{AC}\: \: \: \: ............By\: \: Invertendo\]**

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