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Answers
Answer:
CORRECT QUESTION :
\begin{gathered} \\ \end{gathered}
❖ Prove that if in two triangles, sides of one are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
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SOLUTION :
\begin{gathered} \\ \\ \end{gathered}
❒ Given :-
In two triangles, sides of one are proportional to the sides of the other triangle.
❒ To Prove :-
The corresponding angles of the two triangles are equal and hence the two triangles are similar.
\begin{gathered} \\ \end{gathered}
Suppose,
△ ABC and △ DEF are the two triangles.
Such that,
\sf{\dfrac{AB}{DE}}
DE
AB
= \sf{\dfrac{BC}{EF}}
EF
BC
= \sf{\dfrac{CA}{FD}}
FD
CA
Thus, we have to prove,
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and △ ABC \sim∼ △ DEF
Let us take a point P in DE and Q in FD such that DP = AB and DQ = CA and join PQ.
\begin{gathered} \\ \end{gathered}
Given that,
\sf{\dfrac{AB}{DE}}
DE
AB
= \sf{\dfrac{CA}{FD}}
FD
CA
By construction, we have got,
DP = AB
DQ = CA
∴ \sf{\dfrac{AB}{DE}}
DE
AB
= \sf{\dfrac{CA}{FD}}
FD
CA
⇒ \sf{\dfrac{DP}{DE}}
DE
DP
= \sf{\dfrac{DQ}{FD}}
FD
DQ
⇒ \sf{\dfrac{DE}{DP}}
DP
DE
= \sf{\dfrac{FD}{DQ}}
DQ
FD
⇒ \sf{\dfrac{DE}{DP}}
DP
DE
- 1 = \sf{\dfrac{FD}{DQ}}
DQ
FD
- 1
⇒ \sf{\dfrac{DE - DP}{DP}}
DP
DE−DP
= \sf{\dfrac{FD - DQ}{DQ}}
DQ
FD−DQ
⇒ \sf{\dfrac{PE}{DP}}
DP
PE
= \sf{\dfrac{QF}{DQ}}
DQ
QF
⇒ \sf{\dfrac{DP}{PE}}
PE
DP
= \sf{\dfrac{DQ}{QF}}
QF
DQ
\begin{gathered} \\ \end{gathered}
We know that,
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle.
In △ DEF, PQ line divies DE and FD and \sf{\dfrac{DP}{PE}}
PE
DP
= \sf{\dfrac{DQ}{QF}}
QF
DQ
∴ PQ ॥ EF
Now,
For the lines PQ and EF with transversal PE, we get,
∠P = ∠E (Corresponding Angles) -----------> (1)
And,
For the lines PQ and EF with transversal QF, we get,
∠Q = ∠F (Corresponding Angles) -----------> (2)
Thus, in △ DPQ and △ DEF,
∠P = ∠E [From (1)]
∠Q = ∠F [From (2)]
∴ △ DPQ \sim∼ △ DEF
As the correspondent sides of two similar triangles are proportional, we get,
\sf{\dfrac{DP}{DE}}
DE
DP
= \sf{\dfrac{PQ}{EF}}
EF
PQ
= \sf{\dfrac{DQ}{FD}}
FD
DQ
----------> (3)
Again,
AB = DP
CA = DQ
\sf{\dfrac{AB}{DE}}
DE
AB
= \sf{\dfrac{BC}{EF}}
EF
BC
= \sf{\dfrac{CA}{FD}}
FD
CA
∴ \sf{\dfrac{DP}{DE}}
DE
DP
= \sf{\dfrac{BC}{EF}}
EF
BC
= \sf{\dfrac{DQ}{FD}}
FD
DQ
----------> (4)
\begin{gathered} \\ \end{gathered}
From (4) and (3) we get,
\sf{\dfrac{BC}{EF}}
EF
BC
= \sf{\dfrac{PQ}{EF}}
EF
PQ
∴ BC = PQ ----------> (5)
Now,
In △ ABC and △ DPQ
AB = DP [By construction]
BC= PQ [From (4)]
CA = DQ [By construction]
Hence, by SSS congruency, we get,
△ ABC ≅ △ DPQ
Thus, by CPCT, we get,
∠A = ∠D
∠B = ∠P
∠C = ∠Q
But, from (1) and (2), we get,
∠P = ∠E
∠Q = ∠F
So,
∠B = ∠P = ∠E ----------> (6)
And,
∠C = ∠Q = ∠F ----------> (7)
\begin{gathered} \\ \end{gathered}
Now,
In △ ABC and △ DEF
∠A = ∠D [From CPCT]
∠B = ∠E [From (6)]
∠C = ∠F [From (7)]
Thus, by AAA similarity criteria, we get,
✪ △ ABC \sim∼ △ DEF
Hence proved.
Step-by-step explanation:
1st one
step by step , will send other two in a minute
