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what is the value of x?
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(image of a triangle with points b, q, d on a line; bq=26, qd=39, rc=18, and x is the length of dc, with arrows indicating parallel lines)
Step1: Identify Similar Triangles
Since \( BQ \parallel RC \) (implied by the arrows), triangles \( \triangle BQD \) and \( \triangle RCD \) are similar by the Basic Proportionality Theorem (Thales' theorem). So, the ratios of corresponding sides are equal: \(\frac{BQ}{RC}=\frac{QD}{CD}\). Wait, actually, looking at the segments: \( BQ = 26 \), \( QD = 39 \), \( RC = 18 \), and we need to find \( x = CD \)? Wait, no, maybe \( BR \parallel QC \)? Wait, the diagram: \( BQ = 26 \), \( QD = 39 \), \( RC = 18 \), and we need to find \( x = RD \)? Wait, no, let's re-examine. The triangles: \( B \) to \( Q \) to \( D \), and \( R \) to \( C \) to \( D \), with \( BQ \) and \( RC \) parallel? Wait, the correct proportion: since \( QC \parallel BR \) (arrows indicate parallel), so \( \triangle DQC \sim \triangle DBR \) by AA similarity. Therefore, the ratio of corresponding sides: \( \frac{DQ}{DB}=\frac{DC}{DR} \). Wait, \( DB = BQ + QD = 26 + 39 = 65 \), \( DQ = 39 \), \( DC = x \), \( DR = DC + CR = x + 18 \)? Wait, no, maybe \( RC = 18 \) is \( CR \), and \( x \) is \( CD \). Wait, maybe the correct proportion is \( \frac{BQ}{QD}=\frac{RC}{CD} \)? Wait, no, let's think again. If \( QC \parallel BR \), then by the Basic Proportionality Theorem (Thales' theorem), \( \frac{BQ}{QD}=\frac{RC}{CD} \). Wait, \( BQ = 26 \), \( QD = 39 \), \( RC = 18 \), \( CD = x \). So \( \frac{26}{39}=\frac{18}{x} \)? Wait, no, that would be if \( BQ \) and \( RC \) are corresponding. Wait, maybe \( \frac{BQ}{BD}=\frac{RC}{RD} \). \( BD = 26 + 39 = 65 \), \( RD = x + 18 \). So \( \frac{26}{65}=\frac{18}{x + 18} \). Solving that: \( 26(x + 18) = 65 \times 18 \). \( 26x + 468 = 1170 \). \( 26x = 1170 - 468 = 702 \). \( x = \frac{702}{26} = 27 \). Wait, but maybe the correct proportion is \( \frac{BQ}{QD}=\frac{RC}{CD} \), so \( \frac{26}{39}=\frac{18}{x} \). Then \( 26x = 39 \times 18 \). \( 26x = 702 \). \( x = \frac{702}{26} = 27 \). Yes, that works. So the ratio of \( BQ \) to \( QD \) is equal to the ratio of \( RC \) to \( CD \) because of the parallel lines, so similar triangles. So \( \frac{26}{39}=\frac{18}{x} \). Simplify \( \frac{26}{39} = \frac{2}{3} \). So \( \frac{2}{3}=\frac{18}{x} \). Cross-multiplying: \( 2x = 54 \), so \( x = 27 \). Wait, no, \( 26 \) and \( 39 \) simplify to \( \frac{2}{3} \), so \( \frac{2}{3}=\frac{18}{x} \) would give \( 2x = 54 \), \( x = 27 \). Yes, that's correct.
Step1: Set Up Proportion
Since \( QC \parallel BR \), triangles \( \triangle DQC \) and \( \triangle DBR \) are similar. Thus, \(\frac{BQ}{QD} = \frac{RC}{CD}\). Substituting values: \(\frac{26}{39} = \frac{18}{x}\).
Step2: Simplify and Solve
Simplify \(\frac{26}{39} = \frac{2}{3}\). So \(\frac{2}{3} = \frac{18}{x}\). Cross - multiply: \(2x = 3\times18\). \(2x = 54\). Divide both sides by 2: \(x=\frac{54}{2}=27\).
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