Question

how to prove that ratio of corrosponding sides is the same as the ratio of ccorresponding medians in the similar triangles

Answer 1

$<b> <I>$
Hey there !!

Given :-)


$\bf{ → \triangle ABC \sim \triangle DEF }$


$\bf{ So, \: \angle A = \angle D , \angle B = \angle E , \angle C = \angle F . }$


→ AL and DM are the medians, so BL = CL ; EM = MF.


$\bf{ To \: Prove :-) \frac{BC}{EF} = \frac{AL}{DM} . }$


Proof :-)


$\bf{ We \: have \triangle ABC \sim \triangle DEF . }$


$\bf{ => \frac{AB}{DE} = \frac{BC}{EF} ............(1). }$


$\bf{ => \frac{AB}{DE} = \frac{2BL}{2EM} = \frac{BL}{EM}. }$


$\bf{ => \frac{AB}{DE} = \frac{BL}{EM}. }$


$\bf{ Now, \: in \: \triangle ABL \: and \: \triangle DEM, we \: have }$


$\bf{ => \frac{AB}{DE} = \frac{BL}{EM}. }$
$\bf{ \angle B = \angle E \: (Given). }$


$\bf{ => \triangle ABL \sim \triangle DEM \: [ By \: SAS-similarity ]. }$


$\bf{ => \frac{AB}{DE} = \frac{AL}{DM}. ........(2).}$


▶ From equation (1) and (2), we get


$\huge \boxed{ => \frac{BC}{EF} = \frac{AL}{DM} . }$


✔✔ Hence, it is proved ✅✅.

____________________________________


$\huge \boxed{ \mathbb{THANKS}}$


$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

triangle ABC similar triangle DEF .


=AB/DE = BC/EF ............(1).


=AB/DEBL/2EM= BL/EM
=AB/DE=BL/EM


in triangle ABL and triangle DEM


=AB/DE= BL/EM


angle B = angle E (Given).


ABL similar triangle DEM [ By SAS ]


= AB/DE =AL/DM. ........(2)

equation (1) and (2)

BC/EF = AL/DM