Can anybody send all theorem... Class 10
Answers
Theorem 1
To Prove that, If a line is drawn parallel to one side of a triangle and intersects the other two sides, then the other two sides are divided in the same ratio.
ABC is a triangle, side DE || BC and DE intersects AB at D and AC at E.
Join C to D and B to E. Draw EM ⊥ AC and DN ⊥ AB.
We need to prove that AD/DB = AE/EC
Proof:
Area of a triangle = ½ * AD * EM
Similarly,
Ar(BDE) = ½ * DB * EM
Ar(ADE) = ½* AE * DN
Ar(DEC) = ½ * EC * DN
Hence,
Ar(ADE)/Ar(BDE) = ½ * AD * EM / ½ * DB * EM = AD/DB
Similarly,
Ar(ADE)/Ar(DEC) = AE/EC
Triangles DEC and BDE are on the same base, i.e. DE and between same parallels DE and BC.
Hence,
Ar(BDE) = Ar(DEC)
From above equations, we can say that
AD/DB = AE/EC.
Hence Proved
THEOREM 2
If a line divides any of the two sides of a triangle in the same ratio, then that line is parallel to the third side.
ABC is a triangle in which DE divides AC and AB in the same ratio. This states that:
AB/DB = AE/EC
To prove: If DE` || BC, then
AB/DB = AE`/E`C
According to the theorem AB/DB = AE/EC
Then Accordingly, E and E` must be coincident.
This proves that DE || BC
Proved
THEOREM 3
In two triangles, If the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar, also called SSS (side-side-side) criterion.
Two triangles ABC and DEF are drawn in such a way that their corresponding sides are proportional. It means:
AB/DE = AC/DF = BC/EF
To prove: ∠A = ∠C, ∠B = E and ∠C = ∠F
Hence triangle ABC ~ DEF
In Triangle DEF, draw a line PQ so that DP = AB and DQ = AC
Proof:
because the corresponding sides of the two triangles are equal.
This implies;
DP/PE = DQ/QF = PQ/EF
This also means that ∠P = ∠E and ∠Q = ∠F
We had taken, ∠A=∠D, ∠B=∠P and ∠C=∠Q
Hence; ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
Therefore, from AAA criterion;
Triangle ABC ~ DEF.
Proved