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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) $\triangle APB \cong \triangle AQB$
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) $\triangle APB \cong \triangle AQB$
Solution :
Given Variables & Initial Setup
In $\triangle APB$ and $\triangle AQB$, we are provided with the following information from the geometric configuration:
- Line $l$ is the bisector of $\angle A$, which implies $\angle PAB = \angle QAB$ [By Definition of an Angle Bisector].
- $BP \perp AP$ and $BQ \perp AQ$, which implies $\angle APB = \angle AQB = 90^\circ$ [Given].
- Side $AB$ is common to both triangles [Reflexive Property of Equality].
Visual Representation
Logical Derivation for Congruence
To prove $\triangle APB \cong \triangle AQB$, we evaluate the correspondence between the vertices:
| Condition | Justification |
|---|---|
| $\angle PAB = \angle QAB$ | Since $AB$ is the bisector of $\angle A$. |
| $\angle APB = \angle AQB$ | Both are $90^\circ$ given $BP \perp AP$ and $BQ \perp AQ$. |
| $AB = AB$ | Common side for both triangles. |
Applying the Congruence Criterion
In $\triangle APB$ and $\triangle AQB$:
1. $\angle PAB = \angle QAB$ (Given)
2. $\angle APB = \angle AQB = 90^\circ$ (Given)
3. $AB = AB$ (Common side)
By the AAS (Angle-Angle-Side) Congruence Criterion, since two angles and a non-included side of one triangle are equal to the corresponding two angles and the side of the other triangle, the triangles are congruent.
$\therefore \triangle APB \cong \triangle AQB$ [By AAS Congruence Rule].
Final Solution: $\triangle APB \cong \triangle AQB$ is proven.
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 $AB$.
- 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(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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