CBSE Class 10 Maths (Basic) 2025 · Paper 430/5/1 · Question 31

(i) In $△DFE$ and $△CFB$: $\angle DEF=\angle CBF$ and $\angle EDF=\angle BCF$ (alternate angles, $AE\parallel BC$), so $△DFE\sim △CFB$ by AA. (ii) In $△AEB$ and $△CBF$: $\angle EAB=\angle BCF$ (opposite angles of the parallelogram) and $\angle AEB=\angle CBF$ (alternate angles), so $△AEB\sim △CBF$ by AA. Hence proved.