Question

$\huge \mathscr{\orange {\underline {\pink{\underline{ǫᴜᴇsᴛɪᴏɴ :-}}}}}$


If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Note : All three Cases should be included !!

Quality Answer needed !!

Don't spam plz !!​

Answer 1

Given: Here △ABC and △DEF are such that

∠BAC=∠EDF

ACAB=DFDE

To prove that △ABC∼△DEF

Construction- Draw DB′ equal to AB and DC′ equal to AC in △DEF and join B′C′.

Proof - In △ABC and △DB′C′

AB=BD′ By construction

AC=DC′ by construction

∠A=∠D Given

△ABC≅△DB′C′         ...(S.A.S test of Congruence)

∠B=∠DB′C′         ...C.A.C.T

∠C=∠DC′B′         ....C.A.C.T    

∴△ABC∼△DB′C′          ....(A.A.A test of similarity)

DB′AB=DC′AC=B′C′BC         ....(C.S.S.T)

But DEAB=DFAC       ...Given

Or DEDB′=DFDC′

{AB=DB′,AC=DC′}

⇒B′C′∥EF.

(Side divides the two side in the same ratio then it is parallel to third side).

∴∠DB′C′=∠E=∠B

∠DC′B′=∠F=∠C

∴△ABC∼△DEF    (S.A.S.test of similarity)

Answer 2

Answer:

Construction: Two triangles ABC and DEF are drawn so that their corresponding angles are equal. This means:

∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F

To prove:

theorem on similarity of triangles

Draw a line PQ in the second triangle so that DP = AB and PQ = AC

Proof:

theorem on similarity of triangles

Because corresponding sides of these two triangles are equal

This means; ∠ B = ∠ P = ∠ E and PQ || EF

This means;

theorem on similarity of triangles

Hence;

theorem on similarity of triangles

Hence;

theorem on similarity of triangles

Hrutu id var ye mazya pas deepika la vichr

Step-by-step explanation: