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Q3(i):
Two sides $AB$ and $BC$ and median $AM$ of one triangle ABC are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$ (see Fig. 7.40). Show that: (i) $\triangle ABM \cong \triangle PQN$
Two sides $AB$ and $BC$ and median $AM$ of one triangle ABC are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$ (see Fig. 7.40). Show that: (i) $\triangle ABM \cong \triangle PQN$
Solution :
Given Variables & Initial Setup
In $\triangle ABC$ and $\triangle PQR$, we are given the following conditions:
- $AB = PQ$
- $BC = QR$
- $AM = PN$ (where $AM$ and $PN$ are medians to sides $BC$ and $QR$ respectively).
Step 1: Establishing the Relationship of Sub-segments
By the definition of a median, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $QR$ [Definition of Median].
Therefore:
- $BM = \frac{1}{2}BC$
- $QN = \frac{1}{2}QR$
Since we are given $BC = QR$, it follows by the Euclid's Axiom of substitution that $\frac{1}{2}BC = \frac{1}{2}QR$. Consequently, $BM = QN$.
Step 2: Analytical Proof of Congruence
We now consider the two triangles $\triangle ABM$ and $\triangle PQN$. We evaluate the correspondence of their sides:
| Condition | Justification |
|---|---|
| $AB = PQ$ | Given in problem statement |
| $AM = PN$ | Given in problem statement |
| $BM = QN$ | Derived from $BC=QR$ and midpoint properties |
Step 3: Application of Congruence Criterion
Since all three corresponding sides of $\triangle ABM$ and $\triangle PQN$ are equal, we apply the Side-Side-Side (SSS) Congruence Criterion [SSS Congruence Rule].
$\triangle ABM \cong \triangle PQN$
Final Solution: By the SSS Congruence Criterion, it is proven that $\triangle ABM \cong \triangle PQN$.
More Questions from Class 9 Mathematics Triangles EXERCISE 7.3
- Q1(i): $\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (i) $\triangle ABD \cong \triangle ACD$
- Q1(ii): $\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (ii) $\triangle ABP \cong \triangle ACP$
- Q1(iii): $\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (iii) $AP$ bisects $\angle A$ as well as $\angle D$.
- Q1(iv): $\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (iv) $AP$ is the perpendicular bisector of $BC$.
- Q2(i): $AD$ is an altitude of an isosceles triangle ABC in which $AB = AC$. Show that (i) $AD$ bisects $BC$
- Q2(ii): $AD$ is an altitude of an isosceles triangle ABC in which $AB = AC$. Show that (ii) $AD$ bisects $\angle A$.
- Q3(ii): Two sides $AB$ and $BC$ and median $AM$ of one triangle ABC are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$ (see Fig. 7.40). Show that: (ii) $\triangle ABC \cong \triangle PQR$
- Q4: $BE$ and $CF$ are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
- Q5: ABC is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.
CBSE Solutions for Class 9 Mathematics Triangles
Chapters in CBSE - Class 9 Mathematics
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