NCERT Class 9 Maths Exercise 7.1 Solutions

📘 Concept & Theory Concept Used ›

This question is based on the SAS (Side-Angle-Side) Congruence Rule .

According to the SAS congruence criterion:

If two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle, then the two triangles are congruent.

After proving congruence, we use:

CPCT Principle: Corresponding Parts of Congruent Triangles are Equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the angle bisector property to show: \[\small \angle CAB = \angle BAD \]
Identify the equal sides: \[\small AC = AD \] and \[\small AB = AB \]
Apply SAS congruence criterion.
Use CPCT to conclude: \[\small BC = BD \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.16

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps

Given\[\small AC = AD\]and \(\small AB\) bisects \(\small \angle CAD\).
To Prove\[\small \triangle ABC \cong \triangle ABD\] and \[\small BC = BD\]
Proof
Since \(\small AB\) bisects \(\small \angle CAD\), it divides the angle into two equal parts.
Therefore, \[\small \angle CAB = \angle BAD\]
Now consider triangles \(\small ABC\) and \(\small ABD\).
We have:
First side: \[\small AC = AD\quad\text{Given}\]
Second side: \[\small AB = AB\quad\text{Common Side}\]
Included angle: \[\small\begin{aligned} \angle CAB &^= \angle BAD\\\text{(Because \(AB\)} & \text{ bisects \(\angle CAD\))}\end{aligned}\]
Hence, in triangles \(\small ABC\) and \(\small ABD\),
\[ \begin{aligned} AC &= AD \\ AB &= AB \\ \angle CAB &= \angle BAD \end{aligned} \]
Therefore,
\[\small \triangle ABC \cong \triangle ABD \quad\text{SAS Congruence Rule}\]
Now, by CPCT (Corresponding Parts of Congruent Triangles),
\[\small BC = BD\]
\[\small \triangle ABC \cong \triangle ABD \] and \[\small BC = BD \]

🎯 Exam Significance Exam Significance ›

This is one of the most important introductory questions based on the SAS Congruence Criterion .
Board examinations frequently ask proofs involving:
- Angle bisector property
- Congruent triangles
- Application of CPCT
Competitive entrance exams test the ability to identify:
- Common sides
- Equal angles
- Correct congruence criteria
This problem builds the foundation for advanced geometry proofs in later classes and Olympiad-level reasoning.

📘 Concept & Theory Concept Used ›

This problem is based on the SAS (Side-Angle-Side) Congruence Criterion and the concept of CPCT .

If two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle, then the triangles are congruent.

After proving congruence, we apply:

CPCT: Corresponding Parts of Congruent Triangles are Equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the common side: \[\small AB = BA \]
Use the given equal sides: \[\small AD = BC \]
Use the given equal angles: \[\small \angle DAB = \angle CBA \]
Apply SAS congruence criterion.
Use CPCT to prove: \[\small BD = AC \] and \[\small \angle ABD = \angle BAC \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.17

✏️ Solution Complete Solution ›

Step-by-step Solution · 14 steps

Given\[\small AD = BC\] and \[\small \angle DAB = \angle CBA\]
To Prove
(i) \[\small \triangle ABD \cong \triangle BAC\] (ii) \[\small BD = AC\] (iii)\[\small \angle ABD = \angle BAC\]
Proof
Consider triangles \(\small ABD\) and \(\small BAC\).
In these triangles:
First side: \[\small AD = BC\quad\text{Given}\]
Second side: \[\small AB = BA\quad\text{Common Side}\]
Included angle: \[\small \angle DAB = \angle CBA\quad\text{Given}\]
Therefore,\[\small \begin{aligned} AD &= BC \\ AB &= BA \\ \angle DAB &= \angle CBA \end{aligned}\]
Hence,\[\small\begin{aligned} \triangle ABD &\cong \triangle BAC\\\text{SAS Congru}&\text{ence Criterion}\end{aligned}\]
Now applying CPCT:
Corresponding sides of congruent triangles are equal. Therefore, \[\small BD = AC\]
Also, corresponding angles are equal. Therefore, \[\small \angle ABD = \angle BAC\]
Hence Proved

💡 Answer Final Answer ›

Final Answer:

(i) \(\small \triangle ABD \cong \triangle BAC \) (ii) \(\small BD = AC \) (iii) \(\small \angle ABD = \angle BAC \)

🎯 Exam Significance Exam Significance ›

This is a standard proof-based question from triangle congruence.
Board examinations often ask students to:
- Identify congruent triangles
- Apply SAS criterion correctly
- Use CPCT logically
Competitive exams test whether students can identify corresponding parts quickly and accurately.
This question strengthens geometric reasoning and improves proof-writing skills essential for higher mathematics and Olympiad geometry.

📘 Concept & Theory Concept Used ›

This problem is based on:

Vertically opposite angles
Perpendicular lines forming right angles
AAS (Angle-Angle-Side) Congruence Criterion
CPCT principle

If two angles and one corresponding side of one triangle are equal to those of another triangle, then the two triangles are congruent.

🗺️ Solution Roadmap Step-by-step Plan ›

Let \(\small CD\) intersect \(\small AB\) at point \(\small O\).
Consider triangles: \[\small \triangle DAO \quad \text{and} \quad \triangle CBO \]
Use perpendicular property: \[\small \angle DAO = \angle CBO = 90^\circ \]
Use vertically opposite angles: \[\small \angle DOA = \angle COB \]
Apply AAS congruence criterion.
Use CPCT to prove: \[\small AO = BO \] Therefore, \(CD\) bisects \(AB\).

📊 Graph / Figure Graph / Figure ›

Fig. 7.18

✏️ Solution Complete Solution ›

Step-by-step Solution · 24 steps

Given
\(\small AD\) and \(\small BC\) are equal perpendiculars to line segment \(\small AB\).
Therefore, \[\small AD = BC\] and
\[\small AD \perp AB \quad \text{and} \quad BC \perp AB\]
To Prove
\(\small CD\) bisects \(\small AB\).
That is, \[\small AO = BO\]
where \(\small O\) is the point of intersection of \(CD\) and \(AB\).
Proof
Consider triangles \[\small \triangle DAO \quad \text{and} \quad \triangle CBO \]
Since \(\small AD\) and \(\small BC\) are perpendicular to \(\small AB\), \[\small \angle DAO = 90^\circ\] and \[\small \angle CBO = 90^\circ\]
Therefore, \[\small \angle DAO = \angle CBO\]
Also, lines \(\small CD\) and \(\small AB\) intersect at \(\small O\).
Therefore, vertically opposite angles are equal.
Hence, \[\small \angle DOA = \angle COB\]
Also, \[\small AD = BC\quad\text{Given}\]
Thus, in triangles \(\small DAO\) and \(\small CBO\),
\[ \small \begin{aligned} \angle DAO &= \angle CBO \\ \angle DOA &= \angle COB \\ AD &= BC \end{aligned} \]
Therefore, \[\small \begin{aligned}\triangle DAO &\cong \triangle CBO\\\text{AAS Congru}&\text{ence Criterion}\end{aligned}\]
Applying CPCT:
Corresponding parts of congruent triangles are equal.
Therefore, \[\small AO = BO\]
Hence, point \(\small O\) is the midpoint of \(\small AB\).
\[\small CD \text{ bisects } AB\]
Hence Proved

🎯 Exam Significance Exam Significance ›

This question combines multiple geometry concepts in a single proof.
Board examinations frequently test:
- Perpendicular properties
- Vertically opposite angles
- Triangle congruence
- Use of CPCT
Competitive exams use similar questions to test logical proof-writing and diagram interpretation.
This problem strengthens understanding of midpoint and bisector concepts which are important in coordinate geometry and higher Euclidean geometry.

📘 Concept & Theory Concept Used ›

This question is based on:

Properties of parallel lines
Alternate interior angles
ASA (Angle-Side-Angle) Congruence Criterion

If two angles and the included side of one triangle are equal to the corresponding two angles and included side of another triangle, then the two triangles are congruent.

Since the figure is formed using two pairs of parallel lines, several alternate interior angles become equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Use \[\small l \parallel m \] to identify one pair of alternate interior angles.
Use \[\small p \parallel q \] to identify another pair of alternate interior angles.
Observe that \[\small AC = AC \] because it is the common side.
Apply the ASA congruence criterion.
Conclude: \[\small \triangle ABC \cong \triangle CDA \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.19

✏️ Solution Complete Solution ›

Step-by-step Solution · 13 steps

Given\[\small l \parallel m\] and \[\small p \parallel q\]
To Prove\[\small \triangle ABC \cong \triangle CDA\]
Proof
Consider triangles \[\small \triangle ABC \quad \text{and} \quad \triangle CDA \]
Since \[\small l \parallel m \] and \(AC\) acts as a transversal, alternate interior angles are equal.
Therefore, \[\small \angle BAC = \angle ACD\]
Also, since \[\small p \parallel q \] and \(\small AC\) acts as a transversal, alternate interior angles are equal.
Therefore, \[\small \angle BCA = \angle CAD\]
Further, \[\small AC = AC\] because \(\small AC\) is the common side of both triangles.
Thus, in triangles \(\small ABC\) and \(\small CDA\),
\[\small \begin{aligned} \angle BAC &= \angle ACD \\ AC &= AC \\ \angle BCA &= \angle CAD \end{aligned} \]
Therefore, \[\small \begin{aligned}\triangle ABC &\cong \triangle CDA\\\text{ASA Congru}&\text{ence Criterion}\end{aligned}\]
Hence Proved

🎯 Exam Significance Exam Significance ›

This question develops understanding of parallel line angle properties.
Board examinations frequently include proofs involving:
- Alternate interior angles
- Transversals
- Congruent triangles
Competitive entrance exams test whether students can correctly identify corresponding angles in complex figures.
This problem is also important for advanced geometry, coordinate geometry, and parallelogram-based proofs in higher classes.

Q5

NUMERIC3 marks

Line \(\small l\) is the bisector of angle \(\small \angle A\) and \(\small B\) is any point on \(\small l\). \(\small BP\) and \(\small BQ\) are perpendiculars from \(\small B\) to the arms of \(\small \angle A\). Show that:
(i) \[\small \triangle APB \cong \triangle AQB \]
(ii) \[\small BP = BQ \] or \(\small B\) is equidistant from the arms of \(\small \angle A\).

📘 Concept & Theory Concept Used ›

This question is based on:

Angle bisector property
Perpendicular lines forming right angles
AAS (Angle-Angle-Side) Congruence Criterion
CPCT principle

A point lying on the bisector of an angle is equidistant from the two arms of the angle.

This theorem is proved using congruent triangles.

🗺️ Solution Roadmap Step-by-step Plan ›

Use angle bisector property: \[\small \angle PAB = \angle QAB \]
Since \(\small BP\) and \(\small BQ\) are perpendiculars, identify right angles.
Observe that \[\small AB = AB \] because it is the common side.
Apply the AAS congruence criterion.
Use CPCT to prove: \[\small BP = BQ \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.20

✏️ Solution Complete Solution ›

Step-by-step Solution · 20 steps

Given
Line \(\small l\) bisects \(\small \angle A\).

Point \(\small B\) lies on line \(\small l\).

\(\small BP\) and \(\small BQ\) are perpendiculars drawn from \(\small B\) to the arms of \(\small \angle A\).

Therefore,

\[\small BP \perp AP \]

and

\[\small BQ \perp AQ \]
To Prove
(i) \[\small \triangle APB \cong \triangle AQB \]

(ii) \[\small BP = BQ \]
Proof
Since line \(\small l\) bisects \(\small \angle A\),
\[\small \angle PAB = \angle QAB\]
Also, \[\small BP \perp AP\]
Therefore, \[\small \angle APB = 90^\circ\]
Similarly, \[\small BQ \perp AQ\]
Therefore, \[\small \angle AQB = 90^\circ\]
Hence, \[\small \angle APB = \angle AQB\]
Also, \[\small AB = AB\]
because \(\small AB\) is the common side of both triangles.
Thus, in triangles \(\small APB\) and \(\small AQB\),
\[\small \begin{aligned} \angle PAB &= \angle QAB \\ \angle APB &= \angle AQB \\ AB &= AB \end{aligned} \]
Therefore, \[\small \begin{aligned}\triangle APB &\cong \triangle AQB\\\text{AAS Congru}&\text{ence Criterion}\end{aligned}\]
Applying CPCT:
Corresponding parts of congruent triangles are equal.
Therefore, \[\small BP = BQ\]
Hence, point \(\small B\) is equidistant from the two arms of \(\small \angle A\).
Hence Proved

💡 Answer Final Answer ›

Final Answer:

(i) \( \triangle APB \cong \triangle AQB \) (ii) \( BP = BQ \)

🎯 Exam Significance Exam Significance ›

This question proves an important theorem related to angle bisectors.
Board examinations frequently ask proofs involving:
- Perpendicular distance
- Angle bisectors
- Congruent triangles
- CPCT applications
Competitive entrance exams test whether students can identify right triangles and apply congruence rules correctly.
This theorem is widely used in advanced geometry, coordinate geometry, constructions, and Olympiad problems.

📘 Concept & Theory Concept Used ›

This question is based on:

Angle addition property
SAS (Side-Angle-Side) Congruence Criterion
CPCT principle

If two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle, then the triangles are congruent.

The key idea in this problem is forming equal larger angles by adding the same angle to both equal angles.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the given equal angles: \[\small \angle BAD = \angle EAC \]
Add the common angle \[\small \angle DAC \] to both sides.
Obtain: \[\small \angle BAC = \angle DAE \]
Use: \[\small AB = AD \] and \[\small AC = AE \]
Apply the SAS congruence criterion.
Use CPCT to conclude: \[\small BC = DE \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.21

✏️ Solution Complete Solution ›

Step-by-step Solution · 19 steps

Given \[\small AC = AE\] \[\small AB = AD\] and \[\small \angle BAD = \angle EAC\]
To Prove \[\small BC = DE\]
Proof
Consider triangles \[\small \triangle BAC \quad \text{and} \quad \triangle DAE \]
We are given: \[\small AB = AD\]
and \[\small AC = AE\]
Also, \[\small \angle BAD = \angle EAC\]
Now add \[\small \angle DAC \] to both sides of the above equality.
Then, \[\small \angle BAD + \angle DAC = \angle EAC + \angle DAC\]
Therefore, \[\small \angle BAC = \angle DAE\]
because: \[\small \angle BAC = \angle BAD + \angle DAC\]
and \[\small \angle DAE = \angle DAC + \angle CAE\]
Thus, in triangles \(\small BAC\) and \(\small DAE\),
\[\small \begin{aligned} AB &= AD \\ AC &= AE \\ \angle BAC &= \angle DAE \end{aligned} \]
Therefore, \[\small \begin{aligned}\triangle BAC &\cong \triangle DAE\\\text{SAS Congru}&\text{ence Criterion}\end{aligned}\]
Applying CPCT:
Corresponding parts of congruent triangles are equal.
Therefore,\[\small BC = DE\]
Hence Proved

🎯 Exam Significance Exam Significance ›

This question develops logical understanding of angle addition in geometry proofs.
Board examinations frequently test:
- Formation of larger angles
- Triangle congruence
- Correct use of SAS criterion
- Application of CPCT
Competitive entrance examinations use similar questions to test multi-step geometric reasoning.
This type of proof strengthens analytical thinking and is useful for advanced Euclidean geometry and Olympiad problems.

Q7

NUMERIC3 marks

\(\small AB\) is a line segment and \(\small P\) is its midpoint. \(\small D\) and \(\small E\) are points on the same side of \(\small AB\) such that \[\small \angle BAD = \angle ABE \] and \[\small \angle EPA = \angle DPB \] Show that:
(i) \[\small \triangle DAP \cong \triangle EBP \]
(ii) \[\small AD = BE \]

📘 Concept & Theory Concept Used ›

This problem is based on:

Midpoint property
Angle addition concept
ASA (Angle-Side-Angle) Congruence Criterion
CPCT principle

If two angles and the included side of one triangle are equal to the corresponding two angles and included side of another triangle, then the triangles are congruent.

The important step in this proof is constructing equal larger angles from the given equal angles.

🗺️ Solution Roadmap Step-by-step Plan ›

Use midpoint property: \[\small AP = PB \]
Use the given equal angles: \[\small \angle BAD = \angle ABE \]
Form larger equal angles using: \[\small \angle EPA = \angle DPB \]
Obtain: \[\small \angle DPA = \angle EPB \]
Apply the ASA congruence criterion.
Use CPCT to prove: \[\small AD = BE \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.22

✏️ Solution Complete Solution ›

Step-by-step Solution · 21 steps

Given
\(\small P\) is the midpoint of \(\small AB\).

Therefore,

\[\small AP = PB \]

Also,

\[\small \angle BAD = \angle ABE \]

and

\[\small \angle EPA = \angle DPB \]
To Prove
(i) \[\small \triangle DAP \cong \triangle EBP \]

(ii) \[\small AD = BE \]
Proof
Consider triangles \[\small \triangle DAP \quad \text{and} \quad \triangle EBP \]
Since \(\small A\), \(\small P\), and \(\small B\) are collinear points, the straight line \(\small AB\) forms supplementary angles around \(\small P\).
We are given \[\small \angle EPA = \angle DPB\]
Add the common angle \[\small \angle DPE \] to both sides.
Then, \[\small \angle EPA + \angle DPE = \angle DPB + \angle DPE \]
Therefore, \[\small \angle DPA = \angle EPB \]
Also, \[\small \angle DAP = \angle EBP\]
because: \[\small \angle BAD = \angle ABE\]
and \(\small AP\) and \(\small BP\) lie on the same straight line \(\small AB\).
Further,\[\small AP = PB\]
because \(\small P\) is the midpoint of \(\small AB\).
Thus, in triangles \(\small DAP\) and \(\small EBP\),
\[\small \begin{aligned} \angle DAP &= \angle EBP \\ AP &= PB \\ \angle DPA &= \angle EPB \end{aligned} \]
Therefore, \[\small \begin{aligned}\triangle DAP &\cong \triangle EBP\\\text{ASA Congru}&\text{ence Criterion}\end{aligned}\]
Applying CPCT:
Corresponding parts of congruent triangles are equal.
Therefore, \[\small AD = BE\]
Hence Proved

💡 Answer Final Answer ›

Final Answer:

(i) \( \triangle DAP \cong \triangle EBP \) (ii) \( AD = BE \)

🎯 Exam Significance Exam Significance ›

2-mark standard board question.

Q8

NUMERIC3 marks

In right triangle \(\small ABC\), right angled at \(\small C\), \(\small M\) is the midpoint of hypotenuse \(\small AB\). \(\small C\) is joined to \(\small M\) and produced to a point \(\small D\) such that \[\small DM = CM \] Point \(\small D\) is joined to \(\small B\). Show that:
(i) \[\small \triangle AMC \cong \triangle BMD \]
(ii) \[\small \angle DBC = 90^\circ \]
(iii) \[\small \triangle DBC \cong \triangle ACB \]
(iv) \[\small CM = \frac{1}{2}AB \]

📘 Concept & Theory Concept Used ›

This question combines several important geometry concepts:

SAS Congruence Criterion
Vertically opposite angles
Linear pair angles
RHS Congruence Criterion
Median to hypotenuse theorem
CPCT principle

In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices.

This is one of the most important theorems in coordinate geometry and Euclidean geometry.

🗺️ Solution Roadmap Step-by-step Plan ›

Prove \[\small \triangle AMC \cong \triangle BMD \] using SAS criterion.
Use CPCT to obtain: \[\small AC = BD \]
Use angle relationships to prove: \[\small \angle DBC = 90^\circ \]
Prove \[\small \triangle DBC \cong \triangle ACB \] using RHS criterion.
Finally prove: \[\small CM = \frac{1}{2}AB \]

📊 Graph / Figure Graph / Figure ›

Fig. 7.23

✏️ Solution Complete Solution ›

Step-by-step Solution · 32 steps

Given
\(\small ABC\) is a right triangle right angled at \(\small C\).

Therefore,

\[\small \angle ACB = 90^\circ \]

\(\small M\) is the midpoint of hypotenuse \(\small AB\).

Hence,

\[\small AM = MB \]

Also,

\[\small CM = DM \] (Given)
(i) To Prove \[\small \triangle AMC \cong \triangle BMD\]
Proof
Consider triangles \(\small AMC\) and \(\small BMD\).
We have:\[\small AM = MB \] (Since \(\small M\) is midpoint of \(\small AB\))
\[\small CM = DM\quad\text{Given}\]
Also, \[\small \angle AMC = \angle BMD\]
because they are vertically opposite angles.
Thus, \[\small \begin{aligned} AM &= MB \\ CM &= DM \\ \angle AMC &= \angle BMD \end{aligned} \]
Therefore, \[\small \begin{aligned} \triangle AMC &\cong \triangle BMD\\\text{SAS Congru}&\text{ence Criterion}\end{aligned]\]
(ii) To Prove \[\small \angle DBC = 90^\circ\]
From part (i), by CPCT,
\[\small AC = BD\]
Also, \[\small \angle ACB = 90^\circ\]
Since \(\small A\), \(\small M\), and \(\small B\) are collinear and \(\small C\), \(\small M\), and \(\small D\) are collinear, the corresponding angle structure formed after congruence gives:
\[\small \angle DBC = 90^\circ\]
Therefore, \(\small \triangle DBC\) is right angled at \(\small B\).
(iii) To Prove \[\small \triangle DBC \cong \triangle ACB\]
In triangles \(\small DBC\) and \(\small ACB\),
\[\small \angle DBC = \angle ACB = 90^\circ\]
\[\small DC = AB\]
because: \[\small DC = DM + MC\] and \[\small AB = AM + MB\]
with, \[\small DM = MC \] and \[\small AM = MB \]
Also, \[\small BC = BC \] (Common side)
Therefore,\[\small \begin{aligned} \triangle DBC &\cong \triangle ACB\\\text{RHS Congru}&\text{ence Criterion}\end{aligned}\]
(iv) To Prove \[\small CM = \frac{1}{2}AB\]
Since \(\small M\) is the midpoint of \(\small AB\),
\[\small AM = MB = \frac{1}{2}AB\]
From part (i), by CPCT,\[\small CM = DM\]
and also, \[\small CM = AM\]
Therefore,\[\small CM = \frac{1}{2}AB\]
Hence Proved

🎯 Exam Significance Exam Significance ›

This is one of the most important theorem-based questions from the chapter on triangles.
Board examinations frequently test:
- Midpoint of hypotenuse theorem
- SAS and RHS congruence criteria
- Right triangle properties
- Application of CPCT
Competitive entrance examinations use similar questions to test deep understanding of right triangle geometry.
This theorem is extremely useful in coordinate geometry, trigonometry, circles, and Olympiad geometry problems.

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