Question

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Answer 1

Answer:

This is not the exact ANSWER but a similar one so that you'll get to know how to solve your problem.

Given that:-

△ABC in whichD, E, F are the mid points of sides AB, BC and CA respectively.

To prove:- each of the triangles are similar to the original triangle, i.e.,

△ADF∼△ABC

△BDE∼△ABC

△CEF∼△ABC

Proof:-

Consider the △ADF and △ABC

Since D and F are the mid points of AB and AC respectively.

∴DF∥BC

⇒∠AFD=∠B(Corresponding angles are equal)

Now, in △ADF and △ABC, we have

∠ADF=∠B(Corresponding angles)

∠A=∠A(Common)

By AA similar conditions,

△ADF∼△ABC

Similarly, we have

△BDE∼△ABC

△CEF∼△ABC

∴EF∥AB

⇒EF∥AD..........(1)

And, DE∥AC

⇒DE∥AF..........(2)

From eq

n

(1)&(2), we have

ADEF is a parallelogram.

Similarly, BDFE is a parallelogram.

Now, in △ABC and △DEF

∠A=∠FED(∵Opposite angles of parallelogram)

∠B=∠DFE(∵Opposite angles of parallelogram)

Therefore, by AA similar condition

△ABC∼△DEF

Hence proved that each of the triangles are similar to the original triangle.

Answer 2

Answer:

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