Question

ABCD is a parallelogram. The sides AB and AD are produced to
E and F respectively such that AB = BE and AD = DF. Prove
that triangle BEC is congruent with triangle DCF.​

Answer 1

Given ,

AB // CD & AD // BC

AB = BE

AD = DF

To Prove :-

∆BEC ≈ ∆DEF

Construction :-

AC is joined

Proof :-

Since AF is a straight line therefore,

angle FDC + angle ADC = 180°

=> angle FDC = 90°

Since AE is a straight line therefore,

angle CBE + angle CBA = 180°

=> angle CBE = 90°

therefore, angle FDC = angle CBE

In ∆ADC and ∆ABC

• AD = BC (// sides of parallelogram)
• AB = DC (// sides of parallelogram)
• AC = AC (common)

Therefore, ∆ADC and ∆ABC are congruent by SSS criteria of congruency.

AE = AB + BE

= BE + BE (since AB = BE is given)

= 2BE

AF = AD + DF

AF = AB + DF (since AD and AB are sides of congruent triangle therefore they will be equal)

AF = BE + DF (since AB = BE is given)

Since AD = DF = BE

therefore, DF = BE

In ∆BEC and ∆DEF

• DF = BE (proved above)
• DC = AB (sides of congruent ∆s)
• angle FDC = angle CBE (proved above)

Therefore the triangles are congruent by SAS criteria of congruency.

Hence Proved