Question

PLEASE HELP ME SOLVE THE QUES NO.19

Answer 1

ANSWERIn △DFG and △DAB,wehave ∠1=∠2 [∵AB∣∣DC∣∣EF∴∠1 and ∠2 arecorrespondingangles] ∠FDG=∠ADB [Common]So,byAA−criterionofsimilarity,wehave∴ △DFG∼△DAB⇒ DADF=ABFG.......(i)IntrapeziumABCD,wehave EF∣∣AB∣∣DC∴ DFAF=ECBE⇒ DFAF=43[∵ECBE=43(Given)]⇒ DFAF+1=43+1 [Adding1onbothsides]⇒ DFAF+DF=47⇒ DFAD=47⇒ ADDF=74......(ii)From(i)and(ii),weget ABFG=74⇒ FG=74AB........(iii)In △BEG and △BCD,wehave ∠BEG=∠BCD [Correspondingangles] ∠B=∠B [Common]∴ △BEG∼$\fbox{15 thnx + follow} \blue{ = } \fbox \red{inbox}$

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