Question

If the areas of two similar triangles are equal, prove that they are concurrent.

Answer 1

$Let△lesbe△ABC\&△DEF.$

$Both△lesaresimilar.$

$⟹△ABC\sim △DEF$

$areasareequal.So,ar△ABC=ar△DEF$

$Weknowthatiftwo△lesaresimilar,$

$Ratioofareaisequaltosquareofratio$

$ofitscorrespondingsides.$

$\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}={\left(\frac{BC}{EF}\right)}^2={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AC}{DF}\right)}^2$

$(givenas△ABC=△DEF)$

$1=(\frac{BC}{EF}{)}^2={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AC}{DF}\right)}^2$

$1={\left(\frac{BC}{EF}\right)}^2$

$⟹1=\frac{BC}{EF}$

$\therefore BC=EF$

$1={\left(\frac{AB}{DE}\right)}^2$

$1=\frac{AB}{DE}$

$AB=DE$

$1={\left(\frac{AB}{DE}\right)}^2$

$AC=DF$

$In△ABC\&△DEF$

$EF=BC,AB=DE,AC=DF$

$HencebySSScongruency.$

$△ABC\cong △DEF$

$henceproved$