Triangles
NCERT Solutions • Class 9 Maths • Chapter 7Exercise 7.1
1. In quadrilateral ACBD, $AC = AD$ and $AB$ bisects $\angle A$. Show that $\Delta ABC \cong \Delta ABD$. What can you say about $BC$ and $BD$?
Given: $AC = AD$ and $AB$ bisects $\angle A$ (meaning $\angle CAB = \angle DAB$).
To Prove: $\Delta ABC \cong \Delta ABD$.
Proof: In $\Delta ABC$ and $\Delta ABD$:
1. $AC = AD$ (Given)
2. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
3. $AB = AB$ (Common side)
$\therefore \Delta ABC \cong \Delta ABD$ (By SAS Congruence Rule).
To Prove: $\Delta ABC \cong \Delta ABD$.
Proof: In $\Delta ABC$ and $\Delta ABD$:
1. $AC = AD$ (Given)
2. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
3. $AB = AB$ (Common side)
$\therefore \Delta ABC \cong \Delta ABD$ (By SAS Congruence Rule).
Conclusion about BC and BD:
Since $\Delta ABC \cong \Delta ABD$, their corresponding parts are equal.
$\therefore$ $BC = BD$ (By CPCT – Corresponding Parts of Congruent Triangles).
Since $\Delta ABC \cong \Delta ABD$, their corresponding parts are equal.
$\therefore$ $BC = BD$ (By CPCT – Corresponding Parts of Congruent Triangles).
2. $ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$. Prove that:
(i) $\Delta ABD \cong \Delta BAC$
(ii) $BD = AC$
(iii) $\angle ABD = \angle BAC$
(i) $\Delta ABD \cong \Delta BAC$
(ii) $BD = AC$
(iii) $\angle ABD = \angle BAC$
(i) In $\Delta ABD$ and $\Delta BAC$:
1. $AD = BC$ (Given)
2. $\angle DAB = \angle CBA$ (Given)
3. $AB = AB$ (Common)
$\therefore$ $\Delta ABD \cong \Delta BAC$ (By SAS Rule).
1. $AD = BC$ (Given)
2. $\angle DAB = \angle CBA$ (Given)
3. $AB = AB$ (Common)
$\therefore$ $\Delta ABD \cong \Delta BAC$ (By SAS Rule).
(ii) Since $\Delta ABD \cong \Delta BAC$:
$\therefore$ $BD = AC$ (By CPCT).
$\therefore$ $BD = AC$ (By CPCT).
(iii) Since $\Delta ABD \cong \Delta BAC$:
$\therefore$ $\angle ABD = \angle BAC$ (By CPCT).
$\therefore$ $\angle ABD = \angle BAC$ (By CPCT).
3. $AD$ and $BC$ are equal perpendiculars to a line segment $AB$. Show that $CD$ bisects $AB$.
To Show: $CD$ bisects $AB$ (i.e., $OA = OB$).
Proof: In $\Delta OBC$ and $\Delta OAD$:
1. $\angle BOC = \angle AOD$ (Vertically Opposite Angles)
2. $\angle CBO = \angle DAO = 90^\circ$ (Given perpendiculars)
3. $BC = AD$ (Given)
$\therefore \Delta OBC \cong \Delta OAD$ (By AAS Rule).
Since the triangles are congruent:
$OB = OA$ (By CPCT).
Hence, $CD$ bisects $AB$.
Proof: In $\Delta OBC$ and $\Delta OAD$:
1. $\angle BOC = \angle AOD$ (Vertically Opposite Angles)
2. $\angle CBO = \angle DAO = 90^\circ$ (Given perpendiculars)
3. $BC = AD$ (Given)
$\therefore \Delta OBC \cong \Delta OAD$ (By AAS Rule).
Since the triangles are congruent:
$OB = OA$ (By CPCT).
Hence, $CD$ bisects $AB$.
4. $l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$. Show that $\Delta ABC \cong \Delta CDA$.
Given: $l \parallel m$ and $p \parallel q$.
To Prove: $\Delta ABC \cong \Delta CDA$.
Proof: In $\Delta ABC$ and $\Delta CDA$:
1. $\angle BAC = \angle DCA$ (Alternate Interior Angles, as $p \parallel q$)
2. $\angle BCA = \angle DAC$ (Alternate Interior Angles, as $l \parallel m$)
3. $AC = CA$ (Common side)
$\therefore$ $\Delta ABC \cong \Delta CDA$ (By ASA Rule).
To Prove: $\Delta ABC \cong \Delta CDA$.
Proof: In $\Delta ABC$ and $\Delta CDA$:
1. $\angle BAC = \angle DCA$ (Alternate Interior Angles, as $p \parallel q$)
2. $\angle BCA = \angle DAC$ (Alternate Interior Angles, as $l \parallel m$)
3. $AC = CA$ (Common side)
$\therefore$ $\Delta ABC \cong \Delta CDA$ (By ASA Rule).
5. Line $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$. Show that:
(i) $\Delta APB \cong \Delta AQB$
(ii) $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
(i) $\Delta APB \cong \Delta AQB$
(ii) $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
(i) In $\Delta APB$ and $\Delta AQB$:
1. $\angle APB = \angle AQB = 90^\circ$ (Given perpendiculars)
2. $\angle PAB = \angle QAB$ ($l$ bisects $\angle A$)
3. $AB = AB$ (Common)
$\therefore$ $\Delta APB \cong \Delta AQB$ (By AAS Rule).
1. $\angle APB = \angle AQB = 90^\circ$ (Given perpendiculars)
2. $\angle PAB = \angle QAB$ ($l$ bisects $\angle A$)
3. $AB = AB$ (Common)
$\therefore$ $\Delta APB \cong \Delta AQB$ (By AAS Rule).
(ii) Since $\Delta APB \cong \Delta AQB$:
$\therefore$ $BP = BQ$ (By CPCT).
This means $B$ is equidistant from the arms of $\angle A$.
$\therefore$ $BP = BQ$ (By CPCT).
This means $B$ is equidistant from the arms of $\angle A$.
6. In Fig. 7.21, $AC = AE, AB = AD$ and $\angle BAD = \angle EAC$. Show that $BC = DE$.
Given: $\angle BAD = \angle EAC$.
Add $\angle DAC$ to both sides:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$ … (i)
Add $\angle DAC$ to both sides:
$\angle BAD + \angle DAC = \angle EAC + \angle DAC$
$\Rightarrow \angle BAC = \angle DAE$ … (i)
Proof: In $\Delta ABC$ and $\Delta ADE$:
1. $AB = AD$ (Given)
2. $\angle BAC = \angle DAE$ (From eq i)
3. $AC = AE$ (Given)
$\therefore \Delta ABC \cong \Delta ADE$ (By SAS Rule).
Conclusion: $BC = DE$ (By CPCT).
1. $AB = AD$ (Given)
2. $\angle BAC = \angle DAE$ (From eq i)
3. $AC = AE$ (Given)
$\therefore \Delta ABC \cong \Delta ADE$ (By SAS Rule).
Conclusion: $BC = DE$ (By CPCT).
7. $AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
(i) $\Delta DAP \cong \Delta EBP$
(ii) $AD = BE$
(i) $\Delta DAP \cong \Delta EBP$
(ii) $AD = BE$
Given: $\angle EPA = \angle DPB$.
Add $\angle EPD$ to both sides:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$ … (i)
Add $\angle EPD$ to both sides:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$ … (i)
(i) In $\Delta DAP$ and $\Delta EBP$:
1. $\angle DAP = \angle EBP$ (Given, $\angle BAD = \angle ABE$)
2. $AP = BP$ ($P$ is mid-point of $AB$)
3. $\angle APD = \angle BPE$ (From eq i)
$\therefore$ $\Delta DAP \cong \Delta EBP$ (By ASA Rule).
1. $\angle DAP = \angle EBP$ (Given, $\angle BAD = \angle ABE$)
2. $AP = BP$ ($P$ is mid-point of $AB$)
3. $\angle APD = \angle BPE$ (From eq i)
$\therefore$ $\Delta DAP \cong \Delta EBP$ (By ASA Rule).
(ii) Since $\Delta DAP \cong \Delta EBP$:
$\therefore$ $AD = BE$ (By CPCT).
$\therefore$ $AD = BE$ (By CPCT).
8. In right triangle $ABC$, right angled at $C$, $M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$. Show that:
(i) In $\Delta AMC$ and $\Delta BMD$:
1. $AM = BM$ ($M$ is mid-point)
2. $\angle AMC = \angle BMD$ (Vertically Opposite)
3. $CM = DM$ (Given)
$\therefore$ $\Delta AMC \cong \Delta BMD$ (By SAS Rule).
1. $AM = BM$ ($M$ is mid-point)
2. $\angle AMC = \angle BMD$ (Vertically Opposite)
3. $CM = DM$ (Given)
$\therefore$ $\Delta AMC \cong \Delta BMD$ (By SAS Rule).
(ii) Show $\angle DBC$ is a right angle:
From (i), $\angle MAC = \angle MBD$ (CPCT).
These are alternate interior angles, so $AC \parallel BD$.
Since $AC \parallel BD$, interior angles on the same side sum to $180^\circ$:
$\angle DBC + \angle ACB = 180^\circ$.
$\angle DBC + 90^\circ = 180^\circ \Rightarrow$ $\angle DBC = 90^\circ$.
From (i), $\angle MAC = \angle MBD$ (CPCT).
These are alternate interior angles, so $AC \parallel BD$.
Since $AC \parallel BD$, interior angles on the same side sum to $180^\circ$:
$\angle DBC + \angle ACB = 180^\circ$.
$\angle DBC + 90^\circ = 180^\circ \Rightarrow$ $\angle DBC = 90^\circ$.
(iii) In $\Delta DBC$ and $\Delta ACB$:
1. $DB = AC$ (From (i) CPCT)
2. $\angle DBC = \angle ACB = 90^\circ$ (Proved above)
3. $BC = CB$ (Common)
$\therefore$ $\Delta DBC \cong \Delta ACB$ (By SAS Rule).
1. $DB = AC$ (From (i) CPCT)
2. $\angle DBC = \angle ACB = 90^\circ$ (Proved above)
3. $BC = CB$ (Common)
$\therefore$ $\Delta DBC \cong \Delta ACB$ (By SAS Rule).
(iv) Show $CM = \frac{1}{2} AB$:
From (iii), $DC = AB$ (CPCT).
Given $DM = CM$, so $DC = 2CM$.
$2CM = AB \Rightarrow$ $CM = \frac{1}{2} AB$.
From (iii), $DC = AB$ (CPCT).
Given $DM = CM$, so $DC = 2CM$.
$2CM = AB \Rightarrow$ $CM = \frac{1}{2} AB$.