Question

2 sides AB and BC and the median AM of triangle ABC are respectively equal to side DE and EF and the median DN of triangle DEF .Prove that triangle ABC congruent to triangle DEF.​

Answer 1

________________________


In $ΔABC$, $AM$ is the median to $BC.$


$\bold{Therefore,}$


$BM=BC$


In $ΔDEF$,


$DN$ is the median to $EF.$


$\bold{Therefore,}$


$EN=EF$


$\bold{However,}$


$BC=EF$


$\bold{Therefore,}$


$BC=EF$


$BM=EN$ (1)


In $ΔABM$ and $ΔDEN$


$AB$ = $DE$ (Given)


$BM$ = $EN$ [From equation (1)]


$AM$ = $DN$ (Given)


$\bold{Therefore,}$


$ΔABM$ ≅ $ΔDEN$ (SSS congruence rule)


$ABM$ = $DEN$ (By CPCT)


$ABC$ = $DEF$(2)


$\bold{Now,}$


In $ΔABC$ and $ΔDEF,$


$AB$ = $DE$ (Given)


$ABC$ = $DEF$ [From equation (2)]


$BC$ = $EF$ (Given)


$ΔABC$ ≅ $ΔDEF$ (By SAS congruence rule)


________________________

Answer 2

△ABC and△PQR in which AB=PQ,BC=QR and AM=PN. 

Since AM and PN are median of triangles ABC and PQR respectively. 

Now, BC=QR ∣  Given 

⇒21BC=21QR ∣ Median divides opposite sides in two equal parts

BM=QN... (1) 

Now, in △ABM and△PQN we have 

AB=PQ  ∣ Given

BM=QN ∣  From (i)

and AM=PN ∣  Given

∴ By SSS criterion of congruence, we have 

△ABM≅△PQN, which proves (i) 

∠B=∠Q ... (2)  ∣ Since, corresponding parts of the congruent triangle are equal

Now, in  △ABC and△PQR we have 

AB=PQ ∣ Given

∠B=∠Q ∣ From (2)

BC=QR ∣ Given

∴ by SAS criterion of congruence, we have 

△ABC≅△PQR, which proves (ii)