In Delta ABC,A-P-B,A-Q-C and overline{PQ} par | vedclass

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(B) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two s

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$1$

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(B) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.Since $\overline{PQ} \parallel \overline{BC}$,we have:$\frac{AP}{PB} = \frac{AQ}{QC}$Substituting the given values:$\frac{8x-7}{5x-3} = \frac{4x-3}{3x-1}$Cross-multiplying:$(8x-7)(3x-1) = (4x-3)(5x-3)$$24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9$$24x^2 - 29x + 7 = 20x^2 - 27x + 9$Rearranging the terms to one side:$24x^2 - 20x^2 - 29x + 27x + 7 - 9 = 0$$4x^2 - 2x - 2 = 0$Dividing by $2$:$2x^2 - x - 1 = 0$Factoring the quadratic equation:$2x^2 - 2x + x - 1 = 0$$2x(x-1) + 1(x-1) = 0$$(2x+1)(x-1) = 0$This gives $x = 1$ or $x = -0.5$.Since lengths must be positive,$x$ cannot be $-0.5$ (as $AP = 8(-0.5)-7 = -11$,which is impossible).Therefore,$x = 1$.

Part $I$	Part $II$
$1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$	$a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity.
$2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$	$b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity.
$3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$	$c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity.
$4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$	$d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.

In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $AP = 8x-7$,$PB = 5x-3$,$AQ = 4x-3$ and $QC = 3x-1$,find the value of $x$.

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