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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) $\triangle ABP \cong \triangle ACP$
$\triangle ABC$ 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) $\triangle ABP \cong \triangle ACP$
Solution :
Given Variables & Initial Setup
In $\triangle ABC$ and $\triangle DBC$:
- $AB = AC$ [Given: $\triangle ABC$ is isosceles]
- $DB = DC$ [Given: $\triangle DBC$ is isosceles]
- $AD$ is extended to intersect $BC$ at point $P$.
Step 1: Establishing Triangle Congruence ($\triangle ABD \cong \triangle ACD$)
To prove $\triangle ABP \cong \triangle ACP$, we must first establish properties derived from the larger configuration. Consider $\triangle ABD$ and $\triangle ACD$:
- $AB = AC$ (Given)
- $BD = CD$ (Given)
- $AD = AD$ (Common side)
[By SSS Congruence Criterion, $\triangle ABD \cong \triangle ACD$].
Consequently, by CPCT (Corresponding Parts of Congruent Triangles):
- $\angle BAD = \angle CAD$ (Equation 1)
Step 2: Proving Congruence for $\triangle ABP$ and $\triangle ACP$
Consider $\triangle ABP$ and $\triangle ACP$:
- $AB = AC$ (Given)
- $\angle BAP = \angle CAP$ (From Equation 1, since $P$ lies on the line segment $AD$)
- $AP = AP$ (Common side)
[By SAS Congruence Criterion, $\triangle ABP \cong \triangle ACP$].
Visual Representation of Geometry
Logical Deduction Summary
| Condition | Justification |
|---|---|
| $AB = AC$ | Given isosceles property |
| $\angle BAP = \angle CAP$ | Derived from $\triangle ABD \cong \triangle ACD$ |
| $AP = AP$ | Reflexive property |
Final Solution: Since $AB = AC$, $\angle BAP = \angle CAP$, and $AP = AP$, by the SAS Congruence Criterion, $\triangle ABP \cong \triangle ACP$.
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(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(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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