prove
According to the proportional theorem of common angles , note that $\angle C = \angle C’$, there are:∠C=∠C′,Have:
S△ABCS△A′B′C′=AB⋅ACA′B′⋅A′C′=AB⋅BCA′B′⋅B′C′=AC⋅BCA′C′⋅B′C′thereby
ACA′C′=BCB′C′=ABA′B′Therefore
△ABC△A′B′C′
Theorem 2
(Edge, angle, edge determination theorem) In $\triangle ABC$ and $\triangle A’B’C’$, if $\angle A = \angle A’$, and there is $\frac {AC}{A ‘C’} = \frac {AB}{A’B’}$, then $\triangle ABC \xs \triangle A’B’C’$.△ABCand△A′B′C′in, if∠A=∠A′, and haveACA′C′=ABA′B′,but△ABC△A′B′C′.
prove
Let $\angle A$ coincide with $\angle A’$, write $\frac {AC}{A’C’} = \frac {AB}{A’B’} = k$, then by the proportional theorem of common angles (or triangle area formula) to get∠Aand∠A′coincide, rememberACA′C′=ABA′B′=k, then by
S△ABC′S△ABC=A′C′AC=1k=A′B′AB=S△ACB′S△ABCthereby
S△ABC′=S△ACB′that is
S△BCB′=S△BCC′( parallel line and area relationship theorem ) then $BC \px B’C’$, then ( parallel line property theorem – theorem 3 – equal angles )BCB′C′,but
∠B=∠AB′C′=∠B′,∠C=∠AC′B′=∠C′Therefore
△ABC△A′B′C′
Theorem 3
(Edge, edge, edge determination theorem) In $\triangle ABC$ and $\triangle A’B’C’$, if $\frac {BC}{B’C’} = \frac {AC}{A’ C’} = \frac {AB}{A’B’}$, then $\triangle ABC \xs \triangle A’B’C’$.△ABCand△A′B′C′in, ifBCB′C′=ACA′C′=ABA′B′,but△ABC△A′B′C′.
prove
Might as well set $\frac {AB}{A’B’} = k > 1$, as shown in the figureABA′B′=k>1, as shown in the figure
Take the point $D$ on $AB$ and the point $E$ on $AC$ so that $AD = A’B’$ and $AE = A’C’$. Judging the conditional sides, angles, and sides by the similarity of triangles $\triangle ABC \xs \triangle ADE$, thusABtake pointD,existACtake pointE,MakeAD=A′B′,AE=A′C′. Depend on△ABC△ADE,thus
BCDE=ABAD=ABA′B′=k=BCB′C′Thus $DE = B’C’$, ( the theorem for congruence of triangle angles ) then $\triangle A’B’C’ \qd \triangle ADE$, then $\angle A = \angle A’$, and then Use triangle similarity to determine conditional sides, angles, and sides $\angle ABC \xs \angle A’B’C’$DE=B′C′,△A′B′C′△ADE, then there are∠A=∠A′, then use