Question

Four friends decided to play a game, Meena advised instead of playing physical game let’s play a geometry game. Meena drew a ∆ABC. Rohit found the midpoint of side AB and marked 1 2 it as D. Now the third student Mathew, drew a line DE ‖ BC, for this, D E he made ∠ADE = ∠ABC. Further, the fourth friend Veena om he point E drew a line parallel to AB, she observed that this line cuts BC at new point F. Veena found that EF = BD. As marked in the the given picture. p q r Now all the friends were trying to prove that ∆ADC ≅ ∆EFC.

Answer 1

Answer:

AD=DB (D is the midpoint of AB)

triangle ADC and triangle EFC,

EF = DB (given)

so, AD=EF S

angle ACD = angle ECF (common angle) A

AE = DC

therfore, triangle ADC =~ triangle EFC

Answer 2

Given:

ΔABC, AB has a midpoint D, AC has midpoint E

DE║BC

ADE = ABC

EF = BD

Δ ADE ≅ Δ EFC

To Find

AD = ?

1 and 2 = ?

ADE ≅ EFC ?

AE

Solution:

AD = DB (D is the midpoint of AB)

In ΔADC and ΔEFC,

EF = DB (Given)

so, AD = EF

In Δ ADE and Δ EFC,

∠1 = ∠q

∠ACD = ∠ECF (COMMON ANGLE)

Then, AE = DC

∠2 = ∠r

∴ ΔADC ≅ ΔEFC ( by SAS)

Hence, proved.