Rbse Solutions for Class 9 Maths Chapter 7 Exercise 7.1 | Triangles
1. In quadrilateral ACBD, AC = AD and AB bisect ∠A (see Fig. 7.16). Show that ΔABC≅ ΔABD. What can you say about BC and BD?
Given: $\text{AC} = \text{AD}$, and $\text{AB}$ bisects $\angle \text{CAD}$.
To Prove: $\triangle \text{ABC} \cong \triangle \text{ABD}$.
| Statement | Reason |
| 1. $\text{AC} = \text{AD}$ | Given |
| 2. $\angle \text{CAB} = \angle \text{DAB}$ | Since $\text{AB}$ bisects $\angle \text{A}$ |
| 3. $\text{AB} = \text{AB}$ | Common side |
| 4. $\triangle \text{ABC} \cong \triangle \text{ABD}$ | By SAS (Side-Angle-Side) Congruence Rule |
Conclusion about $\text{BC}$ and $\text{BD}$:
Since $\triangle \text{ABC} \cong \triangle \text{ABD}$, the corresponding parts of congruent triangles are equal.
Therefore, $\mathbf{\text{BC} = \text{BD}}$ (by CPCTC – Corresponding Parts of Congruent Triangles are Congruent).
2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Given: $\text{AD} = \text{BC}$ and $\angle \text{DAB} = \angle \text{CBA}$.
To Prove: $\triangle \text{ABD} \cong \triangle \text{BAC}$.
(i) Prove $\triangle \text{ABD} \cong \triangle \text{BAC}$:
Consider $\triangle \text{ABD}$ and $\triangle \text{BAC}$.
| Statement | Reason |
| 1. $\text{AD} = \text{BC}$ | Given |
| 2. $\angle \text{DAB} = \angle \text{CBA}$ | Given |
| 3. $\text{AB} = \text{BA}$ | Common side |
| 4. $\triangle \text{ABD} \cong \triangle \text{BAC}$ | By SAS (Side-Angle-Side) Congruence Rule |
(ii) Prove $\text{BD} = \text{AC}$:
Since $\triangle \text{ABD} \cong \triangle \text{BAC}$ (from part i), the corresponding sides must be equal.
$$\mathbf{\text{BD} = \text{AC}} \quad \text{(by **CPCTC**)}$$
(iii) Prove $\angle \text{ABD} = \angle \text{BAC}$:
Since $\triangle \text{ABD} \cong \triangle \text{BAC}$ (from part i), the corresponding angles must be equal.
$$\mathbf{\angle \text{ABD} = \angle \text{BAC}} \quad \text{(by **CPCTC**)}$$
3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Given: $\text{AD} = \text{BC}$. $\text{AD} \perp \text{AB}$ and $\text{BC} \perp \text{AB}$. Let $\text{O}$ be the intersection of $\text{AB}$ and $\text{CD}$.
To Prove: $\text{CD}$ bisects $\text{AB}$, i.e., $\text{AO} = \text{BO}$.
Consider $\triangle \text{AOD}$ and $\triangle \text{BOC}$.
| Statement | Reason |
| 1. $\angle \text{DAO} = 90^\circ$ | $\text{AD} \perp \text{AB}$ (Given) |
| 2. $\angle \text{CBO} = 90^\circ$ | $\text{BC} \perp \text{AB}$ (Given) |
| 3. $\angle \text{DAO} = \angle \text{CBO}$ | Both are $90^\circ$ |
| 4. $\angle \text{AOD} = \angle \text{BOC}$ | Vertically Opposite Angles |
| 5. $\text{AD} = \text{BC}$ | Given |
| 6. $\triangle \text{AOD} \cong \triangle \text{BOC}$ | By AAS (Angle-Angle-Side) Congruence Rule |
| 7. $\text{AO} = \text{BO}$ | By CPCTC |
Since $\text{AO} = \text{BO}$, $\mathbf{\text{CD} \text{ bisects } \text{AB}}$.
4. l and m are two parallel lines intersected by
another pair of parallel lines p and q
(see Fig. 7.19). Show that ∆ ABC ≅ ∆ CDA.
Given: $l \parallel m$ and $p \parallel q$. $\text{AC}$ is the transversal.
To Prove: $\triangle \text{ABC} \cong \triangle \text{CDA}$.
Consider $\triangle \text{ABC}$ and $\triangle \text{CDA}$.
| Statement | Reason |
| 1. $\angle \text{BAC} = \angle \text{DCA}$ | $p \parallel q$ and $\text{AC}$ is transversal (Alternate Interior Angles) |
| 2. $\angle \text{BCA} = \angle \text{DAC}$ | $l \parallel m$ and $\text{AC}$ is transversal (Alternate Interior Angles) |
| 3. $\text{AC} = \text{CA}$ | Common side |
| 4. $\triangle \text{ABC} \cong \triangle \text{CDA}$ | By ASA (Angle-Side-Angle) Congruence Rule |
5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Given: $l$ bisects $\angle \text{A}$, $\text{B}$ is on $l$. $\text{BP} \perp \text{AP}$ and $\text{BQ} \perp \text{AQ}$.
To Prove: (i) $\triangle \text{APB} \cong \triangle \text{AQB}$ (ii) $\text{BP} = \text{BQ}$.
(i) Prove $\triangle \text{APB} \cong \triangle \text{AQB}$:
Consider $\triangle \text{APB}$ and $\triangle \text{AQB}$.
| Statement | Reason |
| 1. $\angle \text{QAB} = \angle \text{PAB}$ | $l$ bisects $\angle \text{A}$ (Given) |
| 2. $\angle \text{AQB} = \angle \text{APB}$ | Both are $90^\circ$ ($\text{BP}$ and $\text{BQ}$ are perpendiculars) |
| 3. $\text{AB} = \text{AB}$ | Common side (Hypotenuse) |
| 4. $\triangle \text{APB} \cong \triangle \text{AQB}$ | By AAS (Angle-Angle-Side) Congruence Rule |
(ii) Prove $\text{BP} = \text{BQ}$:
Since $\triangle \text{APB} \cong \triangle \text{AQB}$ (from part i), the corresponding sides must be equal.
$$\mathbf{\text{BP} = \text{BQ}} \quad \text{(by **CPCTC**)}$$
This means that $\mathbf{\text{B}}$ is equidistant from the arms of $\mathbf{\angle \text{A}}$.
6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Given: $\text{AC} = \text{AE}$, $\text{AB} = \text{AD}$, and $\angle \text{BAD} = \angle \text{EAC}$.
To Prove: $\text{BC} = \text{DE}$.
First, we need to show that $\triangle \text{ABC} \cong \triangle \text{ADE}$.
We need $\angle \text{BAC} = \angle \text{DAE}$ for SAS rule.
- $\angle \text{BAD} = \angle \text{EAC}$ (Given)
- Add $\angle \text{DAC}$ to both sides:$$\angle \text{BAD} + \angle \text{DAC} = \angle \text{EAC} + \angle \text{DAC}$$
- This gives:$$\angle \text{BAC} = \angle \text{DAE}$$
Now consider $\triangle \text{ABC}$ and $\triangle \text{ADE}$.
| Statement | Reason |
| 1. $\text{AB} = \text{AD}$ | Given |
| 2. $\angle \text{BAC} = \angle \text{DAE}$ | Proved above |
| 3. $\text{AC} = \text{AE}$ | Given |
| 4. $\triangle \text{ABC} \cong \triangle \text{ADE}$ | By SAS (Side-Angle-Side) Congruence Rule |
| 5. $\mathbf{\text{BC} = \text{DE}}$ | By CPCTC |
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 ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE
Given: $\text{P}$ is the midpoint of $\text{AB}$ ($\implies \text{AP} = \text{BP}$). $\angle \text{BAD} = \angle \text{ABE}$. $\angle \text{EPA} = \angle \text{DPB}$.
To Prove: (i) $\triangle \text{DAP} \cong \triangle \text{EBP}$ (ii) $\text{AD} = \text{BE}$.
(i) Prove $\triangle \text{DAP} \cong \triangle \text{EBP}$:
We need $\angle \text{DPA} = \angle \text{EPB}$ for ASA congruence.
- $\angle \text{EPA} = \angle \text{DPB}$ (Given)
- Add $\angle \text{EPD}$ to both sides:$$\angle \text{EPA} + \angle \text{EPD} = \angle \text{DPB} + \angle \text{EPD}$$
- This gives:$$\angle \text{DPA} = \angle \text{EPB}$$
Now consider $\triangle \text{DAP}$ and $\triangle \text{EBP}$.
| Statement | Reason |
| 1. $\angle \text{PAD} = \angle \text{PBE}$ | $\angle \text{BAD} = \angle \text{ABE}$ (Given) |
| 2. $\text{AP} = \text{BP}$ | $\text{P}$ is the midpoint of $\text{AB}$ |
| 3. $\angle \text{DPA} = \angle \text{EPB}$ | Proved above |
| 4. $\mathbf{\triangle \text{DAP} \cong \triangle \text{EBP}}$ | By ASA (Angle-Side-Angle) Congruence Rule |
(ii) Prove $\text{AD} = \text{BE}$:
Since $\triangle \text{DAP} \cong \triangle \text{EBP}$ (from part i), the corresponding sides must be equal.
$$\mathbf{\text{AD} = \text{BE}} \quad \text{(by **CPCTC**)}$$
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 (see Fig. 7.23). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = ½ AB
Given: $\angle \text{ACB} = 90^\circ$. $\text{M}$ is the midpoint of $\text{AB}$ ($\implies \text{AM} = \text{BM}$). $\text{DM} = \text{CM}$.
To Prove: (i) $\triangle \text{AMC} \cong \triangle \text{BMD}$ (ii) $\angle \text{DBC} = 90^\circ$ (iii) $\triangle \text{DBC} \cong \triangle \text{ACB}$ (iv) $\text{CM} = \frac{1}{2}\text{AB}$.
(i) Prove $\triangle \text{AMC} \cong \triangle \text{BMD}$:
Consider $\triangle \text{AMC}$ and $\triangle \text{BMD}$.
| Statement | Reason |
| 1. $\text{AM} = \text{BM}$ | $\text{M}$ is the midpoint of $\text{AB}$ (Given) |
| 2. $\angle \text{AMC} = \angle \text{BMD}$ | Vertically Opposite Angles |
| 3. $\text{CM} = \text{DM}$ | Given |
| 4. $\mathbf{\triangle \text{AMC} \cong \triangle \text{BMD}}$ | By SAS (Side-Angle-Side) Congruence Rule |
(ii) Prove $\angle \text{DBC}$ is a right angle:
From (i), since $\triangle \text{AMC} \cong \triangle \text{BMD}$:
- $\text{AC} = \text{BD}$ (CPCTC)
- $\angle \text{ACM} = \angle \text{BDM}$ (CPCTC)Since $\angle \text{ACM}$ and $\angle \text{BDM}$ are alternate interior angles for lines $\text{AC}$ and $\text{DB}$ with transversal $\text{CD}$, we conclude that $\mathbf{\text{AC} \parallel \text{DB}}$.Since $\text{AC} \parallel \text{DB}$, $\angle \text{DBC}$ and $\angle \text{ACB}$ are consecutive interior angles (or co-interior angles) for transversal $\text{BC}$.$$\angle \text{DBC} + \angle \text{ACB} = 180^\circ$$We know $\angle \text{ACB} = 90^\circ$:$$\angle \text{DBC} + 90^\circ = 180^\circ$$$$\mathbf{\angle \text{DBC} = 90^\circ}$$
(iii) Prove $\triangle \text{DBC} \cong \triangle \text{ACB}$:
Consider $\triangle \text{DBC}$ and $\triangle \text{ACB}$.
| Statement | Reason |
| 1. $\text{DB} = \text{AC}$ | Proved in (ii) from CPCTC of $\triangle \text{AMC} \cong \triangle \text{BMD}$ |
| 2. $\angle \text{DBC} = \angle \text{ACB}$ | Both are $90^\circ$ (Proved in (ii) and Given) |
| 3. $\text{BC} = \text{CB}$ | Common side |
| 4. $\mathbf{\triangle \text{DBC} \cong \triangle \text{ACB}}$ | By SAS (Side-Angle-Side) Congruence Rule |
(iv) Prove $\text{CM} = \frac{1}{2}\text{AB}$:
From (iii), since $\triangle \text{DBC} \cong \triangle \text{ACB}$:
- $\text{CD} = \text{AB}$ (CPCTC)We know that $\text{CM} = \frac{1}{2}\text{CD}$ (since $\text{DM} = \text{CM}$ and $\text{CD} = \text{CM} + \text{DM}$)Substitute $\text{AB}$ for $\text{CD}$:$$\mathbf{\text{CM} = \frac{1}{2}\text{AB}}$$