Question

Prove this please help anyone
in two triangles the corresponding angles are equal then their corresponding sides are proportional and hence the triangles are similar.​

Answer 1

Step-by-step explanation:

Given:

• Two triangles ∆ABC and ∆DEF.

To Prove:

• ∆ABC ~ ∆DEF

Construction:

• Cut DP = AB and DQ = AC and join PQ

Proof: In these two triangles we are given that corresponding angles are equal.

• ∠A = ∠D
• ∠B = ∠E
• ∠C = ∠F

Now, in ∆ABC and ∆DEF

$\implies{\rm }$ AB = DP {by construction}

$\implies{\rm }$ AC = DQ {by construction}

$\implies{\rm }$ ∠A = ∠D

∴ ∆ABC ≅ ∆DEF by Side angle side criteria of congruence.

Therefore,

➟ ∠B = ∠P and ∠C = ∠Q by CPCT.

➟ ∠P = ∠E { since ∠B = ∠E = ∠P}

But these two angles are corresponding angles so line PQ || EF.

Now by Thales' Theorem we got

$\implies{\rm }$ DP/DE = DQ/DF

$\implies{\rm }$ AB/DE = AC/DF

Similarly,

$\implies{\rm }$ AB/DE = BC/EF

∴ AB/DE = AC/DF = BC/EF

Also,

• ∠A = ∠D
• ∠B = ∠E
• ∠C = ∠F

Hence, ∆ABC ~ ∆DEF

$\large\bold{\texttt {Proved }}$

Answer 2

Given :-

• Two triangles ∆ABC and ∆DEF.

To Prove :-

• ∆ABC ~ ∆DEF

Construction :-

• Cut DP = AB and DQ = AC. Join PQ.

Proof :-

In ∆ABC and ∆DPQ, we have

AB = DP [ by construction ]

AC = DQ [ by construction ]

∠A = ∠D [ given ]

∴ ∆ABC ≈ ∆DPQ [ by SAS-congruence ]

=> ∠B = ∠P

=> ∠E = ∠P [ ∠B = ∠E (given) ]

=> PQ || EF [ corresponding ∠s are equal ]

=> DP/DE = DQ/DF

=> AB/DE = CA/FD [ DP = AB and DQ = AC ]

Similarly,

=> AB/DE = BC/EF

∴ AB/DE = BC/EF = CA/FD

Thus,

• ∠A = ∠D, ∠B = ∠E , ∠C = ∠F and

• AB/DE = BC/EF = CA/FD

Hence,

∆ABC ~ ∆DEF

ProVed

Remark :

Two ∆s are similar <=> they are equiangular.

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