1

Basic Proportionality theorem states - if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.

So in triangle ABC AD/DB = AE/EC if DE line is parallel to BC where D is intersect point at AB and E and AC. But does it also states that -

(a) AD/AB = AE/AC
(b) DB/AB = EC/AC
(c) AD/EC = AE/DB
(d) AD/AE = DB/EC

Can we take this as a property of B.P.T Theorem? Is there any other points that B.P.T can prove and not mentioned here - is ED / BC = AD / AC = AE / EB?

So what are the side ratios which are same?

Programmer
  • 405

1 Answers1

0

enter image description here

(a) comes not from the Basic Proportionality Theorem but similar triangles: $\bigtriangleup ADE\sim \bigtriangleup ABC$ (by AA [angle-angle]) so $\frac{AD}{AB}=\frac{AE}{AC}$.

(b) comes from combining the BPT with (a). We can fuse the identities to get $\frac{AD}{AB}\cdot\frac{DB}{AD}=\frac{AE}{AC}\cdot\frac{EC}{AE}$ or $\frac{DB}{AB}=\frac{EC}{AC}$

(c) doesn't seem to be right. If we cross multiply the BPT, we get $AD\cdot EC=AE\cdot DB$, not $\frac{AD}{EC}=\frac{AE}{DB}$.

(d) comes from rearranging the BPT as well.

By similar triangles we also have $$\frac{ED}{BC}=\frac{AD}{A\color{red}B}=\frac{AE}{\color{orange}{AC}}$$

as well as their rearrangements (one of these cases was found in part a).

Vons
  • 11,004