A point E is taken on the side BC of parallelogram ABCD. AE and DC are

1.	A point E is taken on the side BC of parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (∆ADF) = ar (∆BFC).
Answer» (ABCD) = ar (ABFC) …(i) ar (∆ABF) = ar (∆AFC) ar ∆AFC = \(\frac { 1 }{ 2 }\) ar (∆BCD) …(ii) Also ar (∆ADC) = \(\frac { 1 }{ 2 }\) ar (∆BCD) …(iii) ⇒ ar (∆AFC) = ar (∆ADC) …(iv) ⇒ ar (∆ADF) = ar (∆ADC) + ar (∆ACF) = \(\frac { 1 }{ 2 }\) ar (∆BCD) + \(\frac { 1 }{ 2 }\) ar (∆BFC) [using (ii) and (iii)] = \(\frac { 1 }{ 2 }\) ar (∆BFC) + \(\frac { 1 }{ 2 }\) ar (∆BFC) = ar (∆BFC) ⇒ ar (∆ADF) = ar (∆BFC) Hence proved.

A point E is taken on the side BC of parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (∆ADF) = ar (∆BFC).

Answer»

(ABCD) = ar (ABFC) …(i)

ar (∆ABF) = ar (∆AFC)

ar ∆AFC = \(\frac { 1 }{ 2 }\) ar (∆BCD) …(ii)

Also ar (∆ADC) = \(\frac { 1 }{ 2 }\) ar (∆BCD) …(iii)

⇒ ar (∆AFC) = ar (∆ADC) …(iv)

⇒ ar (∆ADF) = ar (∆ADC) + ar (∆ACF) = \(\frac { 1 }{ 2 }\) ar

(∆BCD) + \(\frac { 1 }{ 2 }\) ar (∆BFC) [using (ii) and (iii)]

= \(\frac { 1 }{ 2 }\) ar (∆BFC) + \(\frac { 1 }{ 2 }\) ar (∆BFC)

= ar (∆BFC)

⇒ ar (∆ADF) = ar (∆BFC)

Hence proved.