Question

9. If mid points of the sides of a triangle are joined, prove that the four triangles so
formed are similar to each other and to the original triangle.

Answer 1

Your Question:

If mid points of the sides of a triangle are joined, prove that the four triangles so  formed are similar to each other and to the original triangle.

Your Answer:

Given:

• The mid points of triangle are joined
• The lines divided the whole triangle into four small triangles

To Prove:

• The triangles formed are similar to each other

or

• △ABC ~ △DEF
• △ABC ~ △ADF
• △ABC ~ △BDE
• △ABC ~ △EFC

Proof:

In △ABC, D, E and F are mid points AB, BC and AC

In △ABC, D and F are mid points AB and AC

So,  DF | | BC   (midpoint theorem)

In △ABC and △ADF  

• ∠A is common  
• ∠ABC = ∠ADF (corresponding angles)

△ABC ~ △ADF (AA similarity) ----------- (1)

Similarly,

• △ABC ~ △BDE (AA similarity) ----------- (2)
• △ABC ~ △EFC (AA similarity) ------------- (3)

In △ABC and △DEF

• DF = BC/2
• DE = AC/2
• EF = AB/2

So, From above

DF/BC = DE/AC = EF/AB

And

△ABC ~ △EFD (SSS similarity)  --------- (4)

From eq 1,2,3 and 4

• △ABC ~ △DEF
• △ABC ~ △ADF
• △ABC ~ △BDF
• △ABC ~ △EFC

Hence Proved

Answer 2

Constructions:

Construct a ∆ABC, where:

• X is the midpoint of AB
• Y is the midpoint of AC
• Z is the midpoint of BC

To Prove:

∆ABC ~ ∆AXY ~ ∆XBZ ~ ∆YZC ~ ∆XYZ

Proof:

X and Y are mid-point of AB and AC respectively, So

➞ XY || BC

➞ ∠A = ∠A ( common )

➞ ∠AYX = ∠ACB ( Corresponding angles )

So, ∆ABC ~ ∆AXY ( By AA ) ─── ( 1 )

X and Z are mid-point of AC and BC respectively, So

➞ XZ || AC

➞ ∠B = ∠B ( common )

➞ ∠BAC = ∠BXZ ( Corresponding angles )

So, ∆ABC ~ ∆XBZ ( By AA ) ─── (2)

Y and Z are mid- point of AC and BC respectively, So

➞ YZ || AB

➞ ∠C = ∠C ( common )

➞ ∠YZC = ∠ABC ( Corresponding angles )

So, ∆YZC ~ ∆ABC ( By AA ) ─── (3)

➞ YZ || AX ( AX is a part of AB ) ... ( a )

➞ XZ || AY ( AY is a part of AC ) ... (b )

From a and b, we can conclude XZYA is a ||gram.

And XY is the diagonal of ||gram XZYA

.

diagonal of parallelogram cut it into two similar triangles So,

∆AXY ~ ∆XYZ ─── (4)

From ( 1 ), (2), (3) and (4)

⟹ ∆ABC ~ ∆AXY ~ ∆XBZ ~ ∆YZC ~∆XYZ

Hence Proved !