91Ó°ÊÓ

In triangles$\mathrm{Δ}DAE$ and $\mathrm{Δ}BCF$.

$\stackrel{→}{AD}â‰\dots \stackrel{→}{BC\text{ â¶Ä‰â€‰}}∴$(ABCD is a rhombus)

$∠ADCâ‰\dots ∠ABC∴$(ABCD is a rhombus)

$∠ADEâ‰\dots ∠CBF\text{ â¶Ä‰}∴$($∠EDC\&∠ABF$ are the equal angle of parallelogram made of parallel sides two parallelograms)

ABCD is a parallelogram.

Similarly,$∠ADCâ‰\dots ∠ABC$ ( ABCD is a rhombus),

$∠EDC\&∠ABF$are equal angles of parallelogram made of parallel sides.

$∠ADEâ‰\dots ∠CBF$

$\stackrel{→}{DE}â‰\dots \stackrel{→}{BF}$$\left(∴BEDF\text{ i²õ²¹â€‰p²¹°ù²¹±ô±ô±ð±ô´Ç²µ°ù²¹³¾}.\right)$

Therefore, by SAS postulate $\mathrm{Δ}DAEâ‰\dots \mathrm{Δ}BCF$.

By CPCT, $\stackrel{→}{AE}â‰\dots \stackrel{→}{FC}$.

In triangles $\mathrm{Δ}AFB\&\mathrm{Δ}CDE$.

$\stackrel{→}{DE}â‰\dots \stackrel{→}{BF}∴\left(BEDF\text{ i²õ²¹±è²¹°ù²¹±ô±ô±ð±ô´Ç²µ°ù²¹³¾}\right)$

Since,$∠EDC\&∠ABF$ are equal angles of parallelogram made of parallel sides.

$∠ADEâ‰\dots ∠CBF\text{ â¶Ä‰}$

$\stackrel{→}{DC}â‰\dots \stackrel{→}{BA}∴\left(BEDF\text{ i²õ²¹±è²¹°ù²¹±ô±ô±ô±ð´Ç²µ°ù²¹³¾}\right)$

Therefore, by SAS postulate $\mathrm{Δ}AFBâ‰\dots \mathrm{Δ}CED$.

By CPCT, $\stackrel{→}{AF}â‰\dots \stackrel{→}{CE}$.

Since two pairs of opposite sides $\stackrel{→}{AE}â‰\dots \stackrel{→}{FC}$and$\stackrel{→}{AF}â‰\dots \stackrel{→}{CE}$ in AECF are parallel, therefore, AECF is a parallelogram.