Question

answer this question to get a brainlist award...First answer deserves it...

Answer 1

Heyy mate ❤✌✌❤

Here's your Answer......

⤵️⤵️⤵️⤵️⤵️⤵️⤵️⤵️

Proof;

F and D are the two mid points on the side AB & AC.

=> DF || BC

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

Thus, in ΔADF and ΔABC, we have

 ∠ADF = ∠B

∠A = ∠A and

∠AFD = ∠B

∴ ΔADF ~ ΔABC

Similarly, we have

ΔBDE ~ ΔABC and ΔCEF ~ ΔABC

Now, we have to show that ΔDEF~ ΔABC.

It can be observed that, ED  II AC and EF  IIAB.

Thus, ADEF is a parallelogram.

∴∠DEF = ∠A  (Opposite angles of a parallelogram are equal)

Similarly, BDFE is a parallelogram.

∴∠DFE = ∠B  (Opposite angles of a parallelogram are equal)

Thus, in ΔDEF and ΔABC, we have

∠DEF = ∠A and ∠DFE = ∠B

Thus, by AA criterion of similarity, we have

ΔDEF~ ΔABC

Thus, each of the triangles ΔADF, ΔBDE,ΔCEF and ΔDEF is similar to ΔABC.
✔✔✔

Answer 2

HEY MATE !!!!!!

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$\huge{Solution}$

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⚫To prove : △ABC ~ △DEF 

△ABC ~ △ADF

△ABC ~ △BDE

△ABC ~ △EFC

⭕Proof: In △ABC, D and E are mid points AB and AC resoopectively.

∴ DF | | BC(midpoint theorem) 

⭕In △ABC = △ADF  

⭕∠A is common;  ∠ADF = ∠ABC (corresponding angles)

⭕△ABC ~ △DF (AA similarity) .......(1) 



⭕Similarly we can prove △ABC ~ △BDE (AA similarity) ..(2)

⭕△ABC ~ △EFC (AA similarity)...(3) 

In △ABC and △DEF;

since D,E,F are the midpoints AB, BC and AC respectively.

DF = (1/2) × BC; DE = (1/2) × AC; EF = (1/2) ×AB; (midpoint theorem)    

∴ AB = BC = CA = 2

EF = DF = DE

∴ △ABC ~ △EFD (SSS similarity)...........(4)

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

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