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Q2(i):
$AD$ is an altitude of an isosceles triangle ABC in which $AB = AC$. Show that (i) $AD$ bisects $BC$
Solution :
Step 1: Given Parameters and Geometric Setup
We are given an isosceles triangle $\triangle ABC$ where $AB = AC$. $AD$ is an altitude drawn from vertex $A$ to the side $BC$. By definition of altitude, $AD \perp BC$, which implies $\angle ADB = \angle ADC = 90^\circ$ [Perpendicularity property].
Step 2: Comparison of Triangles $\triangle ABD$ and $\triangle ACD$
To prove that $D$ is the midpoint of $BC$, we must demonstrate that $BD = CD$ by proving $\triangle ABD \cong \triangle ACD$.
| Condition | Justification |
|---|---|
| $AB = AC$ | Given (Sides of an isosceles triangle) |
| $\angle ADB = \angle ADC = 90^\circ$ | Given ($AD$ is an altitude) |
| $AD = AD$ | Common side [Reflexive Property of Congruence] |
Step 3: Applying the RHS Congruence Criterion
In $\triangle ABD$ and $\triangle ACD$:
- Hypotenuse $AB = AC$ (Given)
- Side $AD = AD$ (Common)
- $\angle ADB = \angle ADC = 90^\circ$ (Altitude property)
Therefore, $\triangle ABD \cong \triangle ACD$ [By RHS Congruence Criterion, which states that if the hypotenuse and one side of a right triangle are equal to the hypotenuse and one side of another right triangle, the triangles are congruent].
Step 4: Conclusion through CPCT
Since the triangles are congruent, their corresponding parts must be equal [By Corresponding Parts of Congruent Triangles (CPCT)].
Thus, $BD = CD$.
Since $D$ lies on $BC$ and $BD = CD$, it follows that $AD$ bisects $BC$.
Final Solution: Since $\triangle ABD \cong \triangle ACD$ by RHS criteria, $BD = CD$ via CPCT, proving that $AD$ bisects $BC$.
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(ii): $AD$ is an altitude of an isosceles triangle ABC in which $AB = AC$. Show that (ii) $AD$ bisects $\angle A$.
- 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) $\triangle ABM \cong \triangle PQN$
- 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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