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Q2(ii):
$AD$ is an altitude of an isosceles triangle ABC in which $AB = AC$. Show that (ii) $AD$ bisects $\angle A$.
Solution :
Given Variables & Initial Setup
In $\triangle ABC$, the following conditions are provided:
- $AB = AC$ (Given, as $\triangle ABC$ is isosceles).
- $AD \perp BC$ (Given, as $AD$ is an altitude, implying $\angle ADB = \angle ADC = 90^\circ$).
Step 1: Identifying Relevant Triangles
To prove that $AD$ bisects $\angle A$ (i.e., $\angle BAD = \angle CAD$), we must establish congruence between $\triangle ABD$ and $\triangle ACD$.
Step 2: Proving Congruence via RHS Criterion
Consider $\triangle ABD$ and $\triangle ACD$:
| Condition | Justification |
|---|---|
| $AB = AC$ | Given (Hypotenuse of both right-angled triangles) |
| $\angle ADB = \angle ADC = 90^\circ$ | Given ($AD$ is an altitude) |
| $AD = AD$ | Common side |
By the RHS (Right-angle, Hypotenuse, Side) Congruence Criterion, we conclude:
$\triangle ABD \cong \triangle ACD$ [By RHS Congruence Rule]
Step 3: Establishing the Bisector Property
Since the triangles are congruent, their corresponding parts must be equal [Per the CPCT - Corresponding Parts of Congruent Triangles theorem]:
$\angle BAD = \angle CAD$
Visual Representation
The equality of $\angle BAD$ and $\angle CAD$ demonstrates that the line segment $AD$ divides the interior angle $\angle BAC$ into two equal parts.
Final Solution: Since $\triangle ABD \cong \triangle ACD$ by the RHS criterion, it follows by CPCT that $\angle BAD = \angle CAD$. Thus, $AD$ bisects $\angle A$.
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 $BC$
- 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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