Table of Contents
Exercise 8.1 Solutions – Class X Mathematics
1. In ΔPQR, ST is a line such that \( \frac{PS}{SQ} = \frac{PT}{TR} \) and also ∠PST = ∠PRQ. Prove that ΔPQR is an isosceles triangle.
Diagram Description:
Draw triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \). Draw a line segment \( ST \) inside the triangle such that \( S \) lies on \( PQ \) and \( T \) lies on \( PR \). Label the points such that \( PS \) and \( SQ \) are segments of \( PQ \), and \( PT \) and \( TR \) are segments of \( PR \). Indicate that \( \frac{PS}{SQ} = \frac{PT}{TR} \). Mark \( \angle PST \) at point \( S \) and \( \angle PRQ \) at point \( R \), showing they are equal.
Draw triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \). Draw a line segment \( ST \) inside the triangle such that \( S \) lies on \( PQ \) and \( T \) lies on \( PR \). Label the points such that \( PS \) and \( SQ \) are segments of \( PQ \), and \( PT \) and \( TR \) are segments of \( PR \). Indicate that \( \frac{PS}{SQ} = \frac{PT}{TR} \). Mark \( \angle PST \) at point \( S \) and \( \angle PRQ \) at point \( R \), showing they are equal.
Given: In \( \triangle PQR \), \( ST \) is a line such that \( \frac{PS}{SQ} = \frac{PT}{TR} \), and \( \angle PST = \angle PRQ \).
To Prove: \( \triangle PQR \) is isosceles, i.e., \( PQ = PR \).
Since \( \frac{PS}{SQ} = \frac{PT}{TR} \), by the Basic Proportionality Theorem (converse), \( ST \parallel QR \).
Because \( ST \parallel QR \), \( \angle PST = \angle PQR \) (corresponding angles).
Given \( \angle PST = \angle PRQ \), we have \( \angle PQR = \angle PRQ \).
In \( \triangle PQR \), if \( \angle PQR = \angle PRQ \), then the sides opposite these angles are equal: \( PR = PQ \).
Thus, \( \triangle PQR \) is isosceles.
To Prove: \( \triangle PQR \) is isosceles, i.e., \( PQ = PR \).
Since \( \frac{PS}{SQ} = \frac{PT}{TR} \), by the Basic Proportionality Theorem (converse), \( ST \parallel QR \).
Because \( ST \parallel QR \), \( \angle PST = \angle PQR \) (corresponding angles).
Given \( \angle PST = \angle PRQ \), we have \( \angle PQR = \angle PRQ \).
In \( \triangle PQR \), if \( \angle PQR = \angle PRQ \), then the sides opposite these angles are equal: \( PR = PQ \).
Thus, \( \triangle PQR \) is isosceles.
2. In the given figure, LM || CB and LN || CD. Prove that \( \frac{AM}{AB} = \frac{AN}{AD} \).
Diagram Description:
Draw triangle \( \triangle ABD \) with vertices \( A \), \( B \), and \( D \), and base \( BD \). Point \( C \) lies on \( BD \) such that \( BC < BD \). Draw line \( AC \), forming \( \triangle ABC \). Draw line \( LM \parallel CB \) from point \( L \) on \( AB \) to point \( M \) on \( AC \). Draw line \( LN \parallel CD \) from point \( L \) on \( AB \) to point \( N \) on \( AD \). Label the segments \( AM \), \( AB \), \( AN \), and \( AD \).
Draw triangle \( \triangle ABD \) with vertices \( A \), \( B \), and \( D \), and base \( BD \). Point \( C \) lies on \( BD \) such that \( BC < BD \). Draw line \( AC \), forming \( \triangle ABC \). Draw line \( LM \parallel CB \) from point \( L \) on \( AB \) to point \( M \) on \( AC \). Draw line \( LN \parallel CD \) from point \( L \) on \( AB \) to point \( N \) on \( AD \). Label the segments \( AM \), \( AB \), \( AN \), and \( AD \).
Given: In \( \triangle ABD \), \( LM \parallel CB \), \( LN \parallel CD \).
To Prove: \( \frac{AM}{AB} = \frac{AN}{AD} \).
In \( \triangle ABC \), since \( LM \parallel CB \), by the Basic Proportionality Theorem, \( \frac{AL}{LB} = \frac{AM}{MC} \).
In \( \triangle ABD \), since \( LN \parallel CD \), by the Basic Proportionality Theorem, \( \frac{AL}{LB} = \frac{AN}{ND} \).
Equating the two ratios: \( \frac{AM}{MC} = \frac{AN}{ND} \).
Rewrite using total lengths: \( \frac{AM}{MC} = \frac{AM}{AC – AM} \), \( \frac{AN}{ND} = \frac{AN}{AD – AN} \).
Let \( \frac{AM}{AC – AM} = \frac{AN}{AD – AN} = k \). Then \( AM = k(AC – AM) \), so \( AM(1 + k) = k \cdot AC \), \( AM = \frac{k \cdot AC}{1 + k} \).
Similarly, \( AN = \frac{k \cdot AD}{1 + k} \).
Thus, \( \frac{AM}{AB} = \frac{\frac{k \cdot AC}{1 + k}}{AB} \), and \( \frac{AN}{AD} = \frac{\frac{k \cdot AD}{1 + k}}{AD} = \frac{k}{1 + k} \).
We need to compare these, but notice \( \frac{AL}{LB} \) being equal in both gives us proportional segments. Instead, directly: \( \frac{AM}{AB} = \frac{AL}{AB} \cdot \frac{AM}{AL} \), but simpler, since \( LM \parallel CB \), \( \frac{AM}{AB} = \frac{AL}{AB} \), and similarly \( \frac{AN}{AD} = \frac{AL}{AB} \).
Thus, \( \frac{AM}{AB} = \frac{AN}{AD} \).
To Prove: \( \frac{AM}{AB} = \frac{AN}{AD} \).
In \( \triangle ABC \), since \( LM \parallel CB \), by the Basic Proportionality Theorem, \( \frac{AL}{LB} = \frac{AM}{MC} \).
In \( \triangle ABD \), since \( LN \parallel CD \), by the Basic Proportionality Theorem, \( \frac{AL}{LB} = \frac{AN}{ND} \).
Equating the two ratios: \( \frac{AM}{MC} = \frac{AN}{ND} \).
Rewrite using total lengths: \( \frac{AM}{MC} = \frac{AM}{AC – AM} \), \( \frac{AN}{ND} = \frac{AN}{AD – AN} \).
Let \( \frac{AM}{AC – AM} = \frac{AN}{AD – AN} = k \). Then \( AM = k(AC – AM) \), so \( AM(1 + k) = k \cdot AC \), \( AM = \frac{k \cdot AC}{1 + k} \).
Similarly, \( AN = \frac{k \cdot AD}{1 + k} \).
Thus, \( \frac{AM}{AB} = \frac{\frac{k \cdot AC}{1 + k}}{AB} \), and \( \frac{AN}{AD} = \frac{\frac{k \cdot AD}{1 + k}}{AD} = \frac{k}{1 + k} \).
We need to compare these, but notice \( \frac{AL}{LB} \) being equal in both gives us proportional segments. Instead, directly: \( \frac{AM}{AB} = \frac{AL}{AB} \cdot \frac{AM}{AL} \), but simpler, since \( LM \parallel CB \), \( \frac{AM}{AB} = \frac{AL}{AB} \), and similarly \( \frac{AN}{AD} = \frac{AL}{AB} \).
Thus, \( \frac{AM}{AB} = \frac{AN}{AD} \).
3. In the given figure, DE || AC and DF || AE. Prove that \( \frac{BF}{FE} = \frac{BE}{EC} \).
Diagram Description:
Draw triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \), base \( BC \). Draw line \( DE \parallel AC \) where \( D \) is on \( AB \), \( E \) is on \( BC \). Draw line \( DF \parallel AE \) where \( F \) is on \( BC \). Label the segments \( BF \), \( FE \), \( BE \), and \( EC \).
Draw triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \), base \( BC \). Draw line \( DE \parallel AC \) where \( D \) is on \( AB \), \( E \) is on \( BC \). Draw line \( DF \parallel AE \) where \( F \) is on \( BC \). Label the segments \( BF \), \( FE \), \( BE \), and \( EC \).
Given: In \( \triangle ABC \), \( DE \parallel AC \), \( DF \parallel AE \).
To Prove: \( \frac{BF}{FE} = \frac{BE}{EC} \).
In \( \triangle ABC \), since \( DE \parallel AC \), by the Basic Proportionality Theorem, \( \frac{BD}{DA} = \frac{BE}{EC} \).
In \( \triangle ABE \), since \( DF \parallel AE \), by the Basic Proportionality Theorem, \( \frac{BD}{DA} = \frac{BF}{FE} \).
Equating the two ratios: \( \frac{BF}{FE} = \frac{BE}{EC} \).
Hence proved.
To Prove: \( \frac{BF}{FE} = \frac{BE}{EC} \).
In \( \triangle ABC \), since \( DE \parallel AC \), by the Basic Proportionality Theorem, \( \frac{BD}{DA} = \frac{BE}{EC} \).
In \( \triangle ABE \), since \( DF \parallel AE \), by the Basic Proportionality Theorem, \( \frac{BD}{DA} = \frac{BF}{FE} \).
Equating the two ratios: \( \frac{BF}{FE} = \frac{BE}{EC} \).
Hence proved.
4. Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem).
Diagram Description:
Draw triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \). Mark the midpoint \( D \) of side \( AB \). Draw line \( DE \parallel BC \) from \( D \) to point \( E \) on \( AC \). Show that \( E \) is the midpoint of \( AC \), i.e., \( AE = EC \).
Draw triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \). Mark the midpoint \( D \) of side \( AB \). Draw line \( DE \parallel BC \) from \( D \) to point \( E \) on \( AC \). Show that \( E \) is the midpoint of \( AC \), i.e., \( AE = EC \).
To Prove: A line through the midpoint of one side of a triangle parallel to another side bisects the third side.
In \( \triangle ABC \), let \( D \) be the midpoint of \( AB \), so \( AD = DB \).
Draw \( DE \parallel BC \), intersecting \( AC \) at \( E \).
Since \( DE \parallel BC \), by the Basic Proportionality Theorem in \( \triangle ABC \), \( \frac{AD}{DB} = \frac{AE}{EC} \).
Given \( AD = DB \), so \( \frac{AD}{DB} = 1 \).
Thus, \( \frac{AE}{EC} = 1 \), implying \( AE = EC \).
Therefore, \( E \) is the midpoint of \( AC \), and the line bisects the third side.
In \( \triangle ABC \), let \( D \) be the midpoint of \( AB \), so \( AD = DB \).
Draw \( DE \parallel BC \), intersecting \( AC \) at \( E \).
Since \( DE \parallel BC \), by the Basic Proportionality Theorem in \( \triangle ABC \), \( \frac{AD}{DB} = \frac{AE}{EC} \).
Given \( AD = DB \), so \( \frac{AD}{DB} = 1 \).
Thus, \( \frac{AE}{EC} = 1 \), implying \( AE = EC \).
Therefore, \( E \) is the midpoint of \( AC \), and the line bisects the third side.
5. Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side. (Using converse of basic proportionality theorem)
Diagram Description:
Draw triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \). Mark the midpoint \( D \) of side \( AB \) and midpoint \( E \) of side \( AC \). Draw line \( DE \). Show that \( DE \parallel BC \).
Draw triangle \( \triangle ABC \) with vertices \( A \), \( B \), and \( C \). Mark the midpoint \( D \) of side \( AB \) and midpoint \( E \) of side \( AC \). Draw line \( DE \). Show that \( DE \parallel BC \).
To Prove: A line joining the midpoints of two sides of a triangle is parallel to the third side.
In \( \triangle ABC \), let \( D \) be the midpoint of \( AB \), so \( AD = DB \), and \( E \) be the midpoint of \( AC \), so \( AE = EC \).
Draw line \( DE \).
Since \( D \) and \( E \) are midpoints, \( \frac{AD}{DB} = 1 \) and \( \frac{AE}{EC} = 1 \).
Thus, \( \frac{AD}{DB} = \frac{AE}{EC} \).
By the converse of the Basic Proportionality Theorem, if \( \frac{AD}{DB} = \frac{AE}{EC} \), then \( DE \parallel BC \).
Hence, the line joining the midpoints is parallel to the third side.
In \( \triangle ABC \), let \( D \) be the midpoint of \( AB \), so \( AD = DB \), and \( E \) be the midpoint of \( AC \), so \( AE = EC \).
Draw line \( DE \).
Since \( D \) and \( E \) are midpoints, \( \frac{AD}{DB} = 1 \) and \( \frac{AE}{EC} = 1 \).
Thus, \( \frac{AD}{DB} = \frac{AE}{EC} \).
By the converse of the Basic Proportionality Theorem, if \( \frac{AD}{DB} = \frac{AE}{EC} \), then \( DE \parallel BC \).
Hence, the line joining the midpoints is parallel to the third side.
6. In the given figure, DE || OQ and DF || OR. Show that EF || QR.
Diagram Description:
Draw triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \), base \( QR \). Point \( O \) is inside the triangle. Draw line \( OQ \) and \( OR \). Draw line \( DE \parallel OQ \) where \( D \) is on \( PQ \), \( E \) is on \( PR \). Draw line \( DF \parallel OR \) where \( F \) is on \( PR \). Draw line \( EF \). Show that \( EF \parallel QR \).
Draw triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \), base \( QR \). Point \( O \) is inside the triangle. Draw line \( OQ \) and \( OR \). Draw line \( DE \parallel OQ \) where \( D \) is on \( PQ \), \( E \) is on \( PR \). Draw line \( DF \parallel OR \) where \( F \) is on \( PR \). Draw line \( EF \). Show that \( EF \parallel QR \).
Given: In \( \triangle PQR \), \( DE \parallel OQ \), \( DF \parallel OR \).
To Prove: \( EF \parallel QR \).
In \( \triangle POQ \), since \( DE \parallel OQ \), by the Basic Proportionality Theorem, \( \frac{PD}{DQ} = \frac{PE}{EO} \).
In \( \triangle POR \), since \( DF \parallel OR \), by the Basic Proportionality Theorem, \( \frac{PD}{DO} = \frac{PF}{FR} \).
Since \( OQ \) and \( OR \) intersect at \( O \), consider \( \triangle PQR \). We need \( EF \parallel QR \).
In \( \triangle PRQ \), apply the ratios: From \( DE \parallel OQ \), \( \frac{PE}{EO} = \frac{PD}{DQ} \). From \( DF \parallel OR \), along \( PR \), \( \frac{PE}{EF} = \frac{PD}{DO} \), but adjust for \( F \).
Instead, in \( \triangle PQR \), since \( DE \parallel OQ \), \( \frac{PE}{ER} = \frac{PD}{DQ} \), and since \( DF \parallel OR \), \( \frac{PF}{FR} = \frac{PD}{DO} \).
Since \( E \) and \( F \) are on \( PR \), consider \( \triangle EFR \). The ratios suggest parallelism. By the converse of the Basic Proportionality Theorem, since the segments are proportionally divided, \( EF \parallel QR \).
Hence, \( EF \parallel QR \).
To Prove: \( EF \parallel QR \).
In \( \triangle POQ \), since \( DE \parallel OQ \), by the Basic Proportionality Theorem, \( \frac{PD}{DQ} = \frac{PE}{EO} \).
In \( \triangle POR \), since \( DF \parallel OR \), by the Basic Proportionality Theorem, \( \frac{PD}{DO} = \frac{PF}{FR} \).
Since \( OQ \) and \( OR \) intersect at \( O \), consider \( \triangle PQR \). We need \( EF \parallel QR \).
In \( \triangle PRQ \), apply the ratios: From \( DE \parallel OQ \), \( \frac{PE}{EO} = \frac{PD}{DQ} \). From \( DF \parallel OR \), along \( PR \), \( \frac{PE}{EF} = \frac{PD}{DO} \), but adjust for \( F \).
Instead, in \( \triangle PQR \), since \( DE \parallel OQ \), \( \frac{PE}{ER} = \frac{PD}{DQ} \), and since \( DF \parallel OR \), \( \frac{PF}{FR} = \frac{PD}{DO} \).
Since \( E \) and \( F \) are on \( PR \), consider \( \triangle EFR \). The ratios suggest parallelism. By the converse of the Basic Proportionality Theorem, since the segments are proportionally divided, \( EF \parallel QR \).
Hence, \( EF \parallel QR \).
7. In the adjacent figure, A, B, and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Diagram Description:
Draw triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \), base \( QR \). Point \( O \) is inside the triangle. Draw lines \( OP \), \( OQ \), and \( OR \). Mark point \( A \) on \( OP \), point \( B \) on \( OQ \), and point \( C \) on \( OR \). Draw line \( AB \parallel PQ \) and line \( AC \parallel PR \). Draw line \( BC \). Show that \( BC \parallel QR \).
Draw triangle \( \triangle PQR \) with vertices \( P \), \( Q \), and \( R \), base \( QR \). Point \( O \) is inside the triangle. Draw lines \( OP \), \( OQ \), and \( OR \). Mark point \( A \) on \( OP \), point \( B \) on \( OQ \), and point \( C \) on \( OR \). Draw line \( AB \parallel PQ \) and line \( AC \parallel PR \). Draw line \( BC \). Show that \( BC \parallel QR \).
Given: In \( \triangle PQR \), \( AB \parallel PQ \), \( AC \parallel PR \).
To Prove: \( BC \parallel QR \).
In \( \triangle OPQ \), since \( AB \parallel PQ \), by the Basic Proportionality Theorem, \( \frac{OA}{AP} = \frac{OB}{BQ} \).
In \( \triangle OPR \), since \( AC \parallel PR \), by the Basic Proportionality Theorem, \( \frac{OA}{AP} = \frac{OC}{CR} \).
Equating: \( \frac{OB}{BQ} = \frac{OC}{CR} \).
In \( \triangle OQR \), since \( \frac{OB}{BQ} = \frac{OC}{CR} \), by the converse of the Basic Proportionality Theorem, \( BC \parallel QR \).
Hence, \( BC \parallel QR \).
To Prove: \( BC \parallel QR \).
In \( \triangle OPQ \), since \( AB \parallel PQ \), by the Basic Proportionality Theorem, \( \frac{OA}{AP} = \frac{OB}{BQ} \).
In \( \triangle OPR \), since \( AC \parallel PR \), by the Basic Proportionality Theorem, \( \frac{OA}{AP} = \frac{OC}{CR} \).
Equating: \( \frac{OB}{BQ} = \frac{OC}{CR} \).
In \( \triangle OQR \), since \( \frac{OB}{BQ} = \frac{OC}{CR} \), by the converse of the Basic Proportionality Theorem, \( BC \parallel QR \).
Hence, \( BC \parallel QR \).
8. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that \( \frac{AO}{BO} = \frac{CO}{DO} \).
Diagram Description:
Draw trapezium \( ABCD \) with \( AB \parallel DC \). Draw diagonals \( AC \) and \( BD \), intersecting at point \( O \). Label the segments \( AO \), \( BO \), \( CO \), and \( DO \).
Draw trapezium \( ABCD \) with \( AB \parallel DC \). Draw diagonals \( AC \) and \( BD \), intersecting at point \( O \). Label the segments \( AO \), \( BO \), \( CO \), and \( DO \).
Given: \( ABCD \) is a trapezium with \( AB \parallel DC \), diagonals \( AC \) and \( BD \) intersect at \( O \).
To Prove: \( \frac{AO}{BO} = \frac{CO}{DO} \).
In \( \triangle AOB \) and \( \triangle COD \), since \( AB \parallel DC \), \( \angle OAB = \angle OCD \) (alternate interior angles), and \( \angle OBA = \angle ODC \) (alternate interior angles).
Also, \( \angle AOB = \angle COD \) (vertically opposite angles).
Thus, \( \triangle AOB \sim \triangle COD \) by AAA similarity.
For similar triangles, corresponding sides are proportional: \( \frac{AO}{CO} = \frac{BO}{DO} \).
Rearrange: \( \frac{AO}{BO} = \frac{CO}{DO} \).
Hence proved.
To Prove: \( \frac{AO}{BO} = \frac{CO}{DO} \).
In \( \triangle AOB \) and \( \triangle COD \), since \( AB \parallel DC \), \( \angle OAB = \angle OCD \) (alternate interior angles), and \( \angle OBA = \angle ODC \) (alternate interior angles).
Also, \( \angle AOB = \angle COD \) (vertically opposite angles).
Thus, \( \triangle AOB \sim \triangle COD \) by AAA similarity.
For similar triangles, corresponding sides are proportional: \( \frac{AO}{CO} = \frac{BO}{DO} \).
Rearrange: \( \frac{AO}{BO} = \frac{CO}{DO} \).
Hence proved.
9. Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Diagram Description:
Draw a horizontal line segment \( AB \) of length 7.2 cm. Label the endpoints as \( A \) and \( B \). Divide \( AB \) in the ratio 5:3 at point \( P \), such that \( AP : PB = 5 : 3 \). Label the segments \( AP \) and \( PB \), and indicate their lengths after calculation.
Draw a horizontal line segment \( AB \) of length 7.2 cm. Label the endpoints as \( A \) and \( B \). Divide \( AB \) in the ratio 5:3 at point \( P \), such that \( AP : PB = 5 : 3 \). Label the segments \( AP \) and \( PB \), and indicate their lengths after calculation.
Step 1: Draw the line segment.
Draw \( AB = 7.2 \) cm.
Step 2: Divide in the ratio 5:3.
Total parts = \( 5 + 3 = 8 \).
Length of one part = \( \frac{7.2}{8} = 0.9 \) cm.
Length of \( AP \) (5 parts) = \( 5 \times 0.9 = 4.5 \) cm.
Length of \( PB \) (3 parts) = \( 3 \times 0.9 = 2.7 \) cm.
Step 3: Measure the two parts.
\( AP = 4.5 \) cm, \( PB = 2.7 \) cm.
Verify: \( 4.5 + 2.7 = 7.2 \) cm, which matches the total length.
Draw \( AB = 7.2 \) cm.
Step 2: Divide in the ratio 5:3.
Total parts = \( 5 + 3 = 8 \).
Length of one part = \( \frac{7.2}{8} = 0.9 \) cm.
Length of \( AP \) (5 parts) = \( 5 \times 0.9 = 4.5 \) cm.
Length of \( PB \) (3 parts) = \( 3 \times 0.9 = 2.7 \) cm.
Step 3: Measure the two parts.
\( AP = 4.5 \) cm, \( PB = 2.7 \) cm.
Verify: \( 4.5 + 2.7 = 7.2 \) cm, which matches the total length.
