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Q3:
$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see Fig. 7.18). Show that $CD$ bisects $AB$.
$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see Fig. 7.18). Show that $CD$ bisects $AB$.
Solution :
Given Variables & Initial Setup
We are provided with a line segment $AB$ and two perpendiculars $AD$ and $BC$ drawn to $AB$ such that $AD = BC$ and $\angle OAD = \angle OBC = 90^\circ$. Let the point of intersection of $CD$ and $AB$ be $O$.
Step 1: Identifying Congruent Triangles
Consider $\triangle OAD$ and $\triangle OBC$. We aim to prove these triangles are congruent to establish a relationship between sides $OA$ and $OB$.
- $\angle OAD = \angle OBC = 90^\circ$ [Given: $AD \perp AB$ and $BC \perp AB$]
- $AD = BC$ [Given]
- $\angle AOD = \angle BOC$ [Vertically Opposite Angles are equal]
Step 2: Applying Congruence Criterion
By the Angle-Angle-Side (AAS) congruence criterion, since two angles and a non-included side are equal:
$\triangle OAD \cong \triangle OBC$ [Per AAS Congruence Rule]
Step 3: Deriving Equality of Segments
Since the triangles are congruent, their corresponding parts are equal by the principle of Congruent Parts of Congruent Triangles (CPCT):
$OA = OB$ [Per CPCT]
Step 4: Conclusion of Bisecting Property
Because point $O$ lies on $AB$ and $OA = OB$, it is logically inferred that point $O$ is the midpoint of $AB$. Therefore, the line segment $CD$ intersects and effectively bisects $AB$ at point $O$.
Final Solution: Since $\triangle OAD \cong \triangle OBC$ by the AAS congruence criterion, we have $OA = OB$ by CPCT, which confirms that $CD$ bisects $AB$.
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) $\angle ABD = \angle BAC$.
- 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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