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Q4:
$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. 7.19). Show that $\triangle ABC \cong \triangle CDA$.
$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. 7.19). Show that $\triangle ABC \cong \triangle CDA$.
Solution :
Given Variables & Initial Setup
Consider the quadrilateral $ABCD$ formed by the intersection of two pairs of parallel lines: line $l \parallel m$ and line $p \parallel q$.
- Line $l \parallel m$
- Line $p \parallel q$
- $A, B, C,$ and $D$ are the points of intersection forming the quadrilateral $ABCD$.
Step 1: Establishing Angle Equality via Parallelism
Since line $l \parallel m$ and line $AC$ acts as a transversal:
- $\angle BAC = \angle DCA$ [By Alternate Interior Angles Theorem, as $l \parallel m$]
Since line $p \parallel q$ and line $AC$ acts as a transversal:
- $\angle BCA = \angle DAC$ [By Alternate Interior Angles Theorem, as $p \parallel q$]
Step 2: Identifying the Common Side
Observe the two triangles $\triangle ABC$ and $\triangle CDA$:
- $AC = AC$ [Common side to both triangles; reflexive property of congruence]
Step 3: Applying Congruence Criterion
We now have two angles and the included side of $\triangle ABC$ equal to two angles and the included side of $\triangle CDA$:
| Condition | Justification |
|---|---|
| $\angle BAC = \angle DCA$ | Alternate Interior Angles |
| $AC = AC$ | Common Side |
| $\angle BCA = \angle DAC$ | Alternate Interior Angles |
By the ASA (Angle-Side-Angle) Congruence Criterion, we conclude that the two triangles are congruent.
Final Solution: Since $\angle BAC = \angle DCA$, $AC = AC$, and $\angle BCA = \angle DAC$, therefore $\triangle ABC \cong \triangle CDA$ by the ASA congruence rule.
More Questions from Class 9 Mathematics Triangles EXERCISE 7.1
- 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 $BD$?
- 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 $AB$.
- 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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