Question

Prove that the line segments joining the midpoints of the sides of a triangle from four triangles each of which is similar to the original triangle.

Answer 1

$△ABC$ in which $D,E,F$ are the mid-points of sides $BC,CA,$and $AB$, respectively. We shall prove that each of the triangles $AFE,FBD,EDC,AND,DEF$ is similar to $△ABC$ $△s$ AFE and ABC

$F$ and $E$ are mid-points of $AB$ and $AC$ respectively

so $FE\parallel BC$

$\Rightarrow \angle AFE=\angle B$ [ Corresponding angles]

In $△ABC$

$\angle AFC=\angle B$

AND $△AFE\sim △ABC$ [By AA corollary ]

Similarly $△FBD\sim △ABC$

and $△EDC\sim △ABC$

Next we shall show that $△DEF\sim △ABC$

clearly $ED\parallel AF\parallel DF\parallel EA$

$\therefore AFDE$ is a parallelogram

$\Rightarrow \angle EDF=\angle A$

$[\because oppositeanglesofa{\parallel }^{gm}areequal]$

Similarly BDEF is a parallelogram

$\Rightarrow \angle DEF=\angle B$

$[\because oppositeanglesofa{\parallel }^{gm}areequal]$

thus in triangles DEF and ABC we have

$\angle EDF=\angle A$

and $\angle DEF=\angle B$

Do by AA corollary

$△DEF\sim △ABC$

Hence each one of the triangle $AFE,FBD,EDC$ and $DEF$ is similar to $△ABC$