Given triangle ABC let X Y and Z be the midpoints of sides B

Given triangle ABC, let X, Y and Z be the midpoints of sides BC,, AC, and AB. Draw triangle XYZ and show that it divides the original triangle into four congruent triangles.

Solution

Note: The line joining the midpoints of two sides of a triangle is parallel to third side and half of it.
Hence YZ = (1/2) BC
(YZ/BC) = (1/2) (1)
Similarly, (XZ/AC) = (1/2) (2)
(XY/AB) = (1/2) (3)
From (1), (2) and (3) we have

YZ/BC=XZ/AC=XY/AB=1/2

But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar

Hence ABC ~ XYZ [By SSS similarity theorem]

It is known that if the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Area(ABC)/Area(XYZ)=YZ^2/BC^2

We know BC=2YZ

So (YZ/BC)^2= (1/2)^2= 1/4

Hence Area(XYZ):Area(ABC)=1:4

Also since YZ=1/2BC and X is Midpoint of BC, YZ=BX=CX

Similarly XY=BZ=AZ and XZ=AY=CY

Let us consider Triangles XYZ and XBZ

XY=BZ, XZ=XZ, BX=YZ

By SSS property BXZ XYZ

Similarly, in triangles XYZ and CXY

XY=XY, XZ=CY, YZ=CX

By SSS property CXY XYZ

In triangles XYZ and AYZ

YZ=YZ, AZ=XY, AY=XZ

By SSS property AYZ XYZ

Thus XYZ AYZ CXY BXZ and

Area(XYZ)=Area(AYZ)=Area( CXY)=Area(BXZ)=1/4(Area(ABC))

Thus XYZ divides ABC into 4 congruent triangles.

Hence Proved