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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 $BC$.
$\triangle ABC$ 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 $BC$.
Solution :
Step 1: Establishing Geometric Preliminaries
Given that $\triangle ABC$ is isosceles with $AB = AC$ and $\triangle DBC$ is isosceles with $DB = DC$. We must prove that $AP \perp BC$ and $BP = PC$, implying $AP$ is the perpendicular bisector of $BC$.
Step 2: Proving Congruency of Side Triangles
Consider $\triangle ABD$ and $\triangle ACD$:
- $AB = AC$ (Given, as $\triangle ABC$ is isosceles)
- $BD = CD$ (Given, as $\triangle DBC$ is isosceles)
- $AD = AD$ (Common side)
[By SSS Congruence Criterion, $\triangle ABD \cong \triangle ACD$].
From the congruency above, by Corresponding Parts of Congruent Triangles (CPCT), we establish:
$\angle BAD = \angle CAD$ (Equation 1)
Step 3: Analyzing $\triangle ABP$ and $\triangle ACP$
Now consider $\triangle ABP$ and $\triangle ACP$:
- $AB = AC$ (Given)
- $\angle BAP = \angle CAP$ (From Equation 1, since $P$ lies on $AD$)
- $AP = AP$ (Common side)
[By SAS Congruence Criterion, $\triangle ABP \cong \triangle ACP$].
Step 4: Deriving the Perpendicular Bisector Property
From the congruency $\triangle ABP \cong \triangle ACP$, we obtain the following results via CPCT:
1. $BP = CP$ [This confirms $P$ is the midpoint of $BC$].
2. $\angle APB = \angle APC$ [Corresponding angles].
Since $\angle APB + \angle APC = 180^\circ$ [Linear Pair Axiom], and $\angle APB = \angle APC$, it follows that:
$2 \angle APB = 180^\circ \implies \angle APB = 90^\circ$.
Step 5: Conclusion
Since $\angle APB = 90^\circ$, $AP \perp BC$. Combined with $BP = PC$, it is analytically demonstrated that line segment $AP$ bisects $BC$ at a $90^\circ$ angle.
Final Solution: $AP$ is the perpendicular bisector of $BC$.
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 $\angle D$.
- 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(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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