Question

prove that triangle ABE and triangle ACD are congruent​

Answer 1

From the diagram

i)

BC // DE

Area of ∆CBE = Area of ∆CBD ===> eqn 1

ii)

Area of ∆ABE = Ar ∆ABC + Ar ∆CBE

==> Ar ∆ABC + Ar ∆CBD [ :. from eqn 1 ]

==> Ar ∆ACD

Area of ∆ABE = Area of ∆ACD

Hence proved.

$\huge\bold\pink{Ms~Coward}$

Answer 2

Hi! :D

Thanks for the question.

Given :

• In triangle AED, BC is parallel to DE.

To prove :

• A (Δ ABE) = A (ΔACD)

Proof :

Let's first consider the area of triangle BCE,

$\tt{A(\triangle\:BCE)\:=\:{\dfrac{1}{2}\:\times\:base\:\times\:height\:}}$

$\tt{A(\triangle\:BCE)\:=\:{\dfrac{1}{2}\:\times\:BC\:\times\:BE\:---(i)}}$

Now, consider the area of triangle CBD,

$\tt{A(\triangle\:CBD)\:=\:{\dfrac{1}{2}\:\times\:base\:\times\:height\:}}$

$\tt{A(\triangle\:CBD)\:=\:{\dfrac{1}{2}\:\times\:BC\:\times\:BE\:---(ii)}}$

From (i) and (ii),

$\tt{A(\triangle\:CBD)\:= A(\triangle\:BCE)---(iii)}$

Area of larger triangle, ABE would be sum of area of triangle ABC & area of triangle BCE.

$\tt{A(\triangle\:ABE)\:=\:A(\triangle\:ABC)\:+\:A(\triangle\:BCE)}$

From (iii) we can replace BCE with CBD.

$\tt{A(\triangle\:ABE)\:=\:A(\triangle\:ABC)\:+\:A(\triangle\:CBD)}$

$\tt{A(\triangle\:ABE)\:=\:A(\triangle\:ACD)}$$\underbrace{\sf{\triangle\:ABC\:and\:\triangle\:CBD\:adds\:up\:to\:form\:\triangle\:ACD)}}$

Hence proved!