Question
In figure, ABCD, ABFE and CDEF are parallelograms. Prove that $\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{BCF}).$
✓
Answer
Given that: ABCD is parallelogram ⇒ AD = BC CDEF is parallelogram ⇒ DE = CF ABFE is parallelogram ⇒ AE = BF Thus, in $\triangle\text{ADE}$ and $\triangle\text{BCF},$ we have AD = BC, DE = CF and AE = BF So, by SSS criterion of congruence, we have$\triangle\text{ADE}\cong\triangle\text{BCF}$
$\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{BCF})$
$\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{BCF})$
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