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Q4:
$BE$ and $CF$ are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution :
Given Variables & Initial Setup
Consider a triangle $ABC$ where:
- $BE \perp AC$ (Altitude from $B$ to side $AC$)
- $CF \perp AB$ (Altitude from $C$ to side $AB$)
- $BE = CF$ (Given equal altitudes)
Step 1: Identification of Triangles
To prove $AB = AC$, we examine triangles $\triangle BCF$ and $\triangle CBE$.
- In $\triangle BCF$ and $\triangle CBE$:
- $\angle BFC = \angle CEB = 90^\circ$ [Per definition of altitudes $BE$ and $CF$]
- $BC = CB$ [Common side for both triangles]
- $CF = BE$ [Given in the problem statement]
Step 2: Applying RHS Congruence Criterion
The Right-Angle Hypotenuse-Side (RHS) Congruence Rule states that if the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle, the two triangles are congruent.
Since:
- $Hypotenuse_{BCF} = Hypotenuse_{CBE} = BC$
- $Side_{CF} = Side_{BE}$
- $\angle BFC = \angle CEB = 90^\circ$
Therefore, $\triangle BCF \cong \triangle CBE$ [By RHS Congruence Rule].
Step 3: Deriving Interior Angles
From the congruence of triangles $\triangle BCF \cong \triangle CBE$, we conclude by Corresponding Parts of Congruent Triangles (CPCT):
$\angle FCB = \angle EBC$
Equivalently, $\angle ACB = \angle ABC$.
Step 4: Proving Triangle ABC is Isosceles
In $\triangle ABC$, we have established that $\angle ABC = \angle ACB$.
According to the theorem: "Sides opposite to equal angles of a triangle are equal."
Since $\angle ABC = \angle ACB$, the side opposite to $\angle ABC$ (which is $AC$) must be equal to the side opposite to $\angle ACB$ (which is $AB$).
$AB = AC$
Final Solution: Since the two sides $AB$ and $AC$ are equal, $\triangle ABC$ is proved to be an isosceles triangle.
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(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) $\triangle ABC \cong \triangle PQR$
- 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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