Question

prove that if two Triangles are equiangular then their corresponding sides are proportional .this is theorem

Answer 1

Answer:

Step-by-step explanation:

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. This is also called AAA (Angle-Angle-Angle) criterion.

theorem on similarity of triangles

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:

A

B

D

E

=

A

C

D

F

=

B

C

E

F

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

Proof:

Δ

A

B

C

≅

Δ

D

P

Q

Because corresponding sides of these two triangles are equal

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

This means;

D

P

P

E

=

D

Q

Q

F

Hence;

A

B

D

E

=

A

C

D

F

A

B

D

E

=

B

C

E

F

Hence;

A

B

D

E

=

A

C

D

F

=

B

C

E

F

proved

Please mark it as brainlist

Answer 2

Given: ∠BAC=∠EDF

∠ABC=∠DEF

To prove: $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$

Construction: Mark points G and H on the side AB and AC

such that

AG=DE , AH=DF

proof: in triangle AGH and DEF

AG=DE.....by construction

AH=DF ..... by contsruction

∠GAH=∠EDF...Given

therefore ,

$\bigtriangleup AGH \cong \bigtriangleup FED$

by SAS congruency

thus

∠AGH=∠DEF  ....by CPCT

but

∠ABC=∠DEF

∠AGH=∠ABC

thus

GH║BC

now , In triangle ABC

$\frac{AB}{AG}=\frac{BC}{GH}=\frac{CA}{HA}$

Hence ,

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$

hence proved .