Question

-State Basic Proportionality Theorem and prove it​

Answer 1

Answer:

Basic Proportionality Theorem (BPT) If a side is parallel to one side of a triangle and it intersects the other two points in two distinct points, the it divides the other two sides in proportion

Answer 2

Answer:

Basic Proportionality Theorem states that, 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.

⇒ Given: In triangle ABC, DE ║ BC

⇒ To prove: $\frac{AD}{DB}$ = $\frac{AE}{EC}$

⇒ Construction: Join BE and CD and draw DG ⊥ AC and EF ⊥ AB.

⇒ Proof:

   ⇒ Area(triangle ADE) = $\frac{1}{2}$ * base * height

                                        = $\frac{1}{2}$  * AD * EF ---------------------(i)

   ⇒ Area(triangle BDE) = $\frac{1}{2}$ * base * height

                                       = $\frac{1}{2}$ * DB * EF -----------------------(ii)

   By Dividing (i) & (ii),

   ⇒ $\frac{Area(ADE)}{Area(BDE)}$ = $\frac{1/2*AD*EF}{1/2*DB*EF}$

   ⇒ $\frac{Ar(ADE)}{Ar(BDE)}$ = $\frac{AD}{DB}$ ------------------------ (a)

   ⇒ Area(triangle ADE) = $\frac{1}{2}$  * AE * DM ---------------------(iii)

   ⇒ Area(triangle DEC) = $\frac{1}{2}$  * EC * DM ---------------------(iv)

   By Dividing (iii) & (iv),

   ⇒ $\frac{Area(ADE)}{Area(DEC)}$ = $\frac{1/2*AE*DM}{1/2*EC*DM}$

   ⇒  $\frac{Ar(ADE)}{Ar(DEC)}$ = $\frac{AE}{EC}$ ------------------------ (b)

Now,

Triangle BDE & Triangle DEC are on same base DE and between the same parallel lines BC and DE

∴ Ar(BDE) = Ar(DEC)

Hence,

⇒ $\frac{Area(ADE)}{Area(BDE)}$ = $\frac{Area(ADE)}{Area(DEC)}$

⇒ $\frac{AD}{DB}$ = $\frac{AE}{EC}$       (from (a) and (b))

Hence Proved

HOPE IT HELPS YOU!!!