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Q1:
In quadrilateral ACBD, $AC = AD$ and $AB$ bisects $\angle A$ (see Fig. 7.16). Show that $\triangle ABC \cong \triangle ABD$. What can you say about $BC$ and $BD$?
In quadrilateral ACBD, $AC = AD$ and $AB$ bisects $\angle A$ (see Fig. 7.16). Show that $\triangle ABC \cong \triangle ABD$. What can you say about $BC$ and $BD$?
Solution :
Step 1: Identifying Given Parameters
In the quadrilateral $ACBD$, the following conditions are explicitly stated:
- $AC = AD$ [Given as a side equality]
- $AB$ bisects $\angle CAB$ and $\angle DAB$, implying $\angle CAB = \angle DAB$ [By definition of an angle bisector]
Step 2: Establishing Congruence Criteria
To prove $\triangle ABC \cong \triangle ABD$, we compare the two triangles based on the provided information:
| Condition | Justification |
|---|---|
| $AC = AD$ | Given |
| $\angle CAB = \angle DAB$ | Since $AB$ is the bisector of $\angle CAD$ |
| $AB = AB$ | Common side in both triangles [Reflexive Property of Congruence] |
Step 3: Logical Derivation of Congruence
By observing the two triangles $\triangle ABC$ and $\triangle ABD$:
- Side $AC$ of $\triangle ABC$ is equal to side $AD$ of $\triangle ABD$.
- Included angle $\angle CAB$ of $\triangle ABC$ is equal to included angle $\angle DAB$ of $\triangle ABD$.
- Side $AB$ is common to both.
Therefore, by the SAS (Side-Angle-Side) Congruence Criterion, we conclude:
$\triangle ABC \cong \triangle ABD$
Step 4: Analysis of $BC$ and $BD$
Since $\triangle ABC \cong \triangle ABD$, all corresponding parts of congruent triangles are equal [By CPCT - Corresponding Parts of Congruent Triangles].
Consequently, $BC = BD$.
Step 5: Visual Representation
Final Solution: The triangles $\triangle ABC$ and $\triangle ABD$ are congruent by the SAS congruence criterion. Because they are congruent, it follows that $BC = BD$ by CPCT.
More Questions from Class 9 Mathematics Triangles EXERCISE 7.1
- Q2(i): ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig. 7.17). Prove that (i) $\triangle ABD \cong \triangle BAC$
- Q2(ii): ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig. 7.17). Prove that (ii) $BD = AC$
- Q2(iii): ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig. 7.17). Prove that (iii) $\angle ABD = \angle BAC$.
- Q3: $AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see Fig. 7.18). Show that $CD$ bisects $AB$.
- Q4: $l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. 7.19). Show that $\triangle ABC \cong \triangle CDA$.
- Q5(i): 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 $\angle A$ (see Fig. 7.20). Show that: (i) $\triangle APB \cong \triangle AQB$
- Q5(ii): 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 $\angle A$ (see Fig. 7.20). Show that: (ii) $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
- Q6: In Fig. 7.21, $AC = AE$, $AB = AD$ and $\angle BAD = \angle EAC$. Show that $BC = DE$.
- Q7(i): $AB$ 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 $\angle EPA = \angle DPB$ (see Fig. 7.22). Show that (i) $\triangle DAP \cong \triangle EBP$
- Q7(ii): $AB$ 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 $\angle EPA = \angle DPB$ (see Fig. 7.22). Show that (ii) $AD = BE$
- Q8(i): 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) $\triangle AMC \cong \triangle BMD$
- Q8(ii): 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: (ii) $\angle DBC$ is a right angle.
- Q8(iii): 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: (iii) $\triangle DBC \cong \triangle ACB$
- Q8(iv): 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: (iv) $CM = \frac{1}{2} AB$
CBSE Solutions for Class 9 Mathematics Triangles
Chapters in CBSE - Class 9 Mathematics
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