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Q6:
$\triangle ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (see Fig. 7.34). Show that $\angle BCD$ is a right angle.
$\triangle ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (see Fig. 7.34). Show that $\angle BCD$ is a right angle.
Solution :
Given Variables & Initial Setup
We are provided with $\triangle ABC$ such that $AB = AC$. Side $BA$ is extended to $D$ such that $AD = AB$. We aim to prove that $\angle BCD = 90^\circ$.
Step 1: Establishing Triangle Properties
In $\triangle ABC$, given $AB = AC$.
Per the Isosceles Triangle Theorem [Angles opposite to equal sides are equal], we have:
$\angle ACB = \angle ABC$ --- (Equation 1)
Step 2: Utilizing Transitive Property of Equality
We are given $AD = AB$. Since $AB = AC$, by the property of substitution:
$AD = AC$ --- (Equation 2)
Now consider $\triangle ACD$. Since $AD = AC$, $\triangle ACD$ is also isosceles.
Per the Isosceles Triangle Theorem applied to $\triangle ACD$:
$\angle ADC = \angle ACD$ --- (Equation 3)
Step 3: Summation of Interior Angles
Consider the full triangle $\triangle BCD$. The sum of the interior angles is $180^\circ$ [Angle Sum Property of a Triangle]:
$\angle D + \angle CBD + \angle BCD = 180^\circ$
Note that $\angle CBD$ is the same as $\angle ABC$ (as $D, A, B$ are collinear). Also, $\angle BCD$ is composed of $\angle BCA + \angle ACD$.
Substituting values:
$\angle ADC + \angle ABC + (\angle BCA + \angle ACD) = 180^\circ$
Step 4: Algebraic Substitution
From Equation 1 and Equation 3, we substitute variables:
Since $\angle ADC = \angle ACD$ and $\angle ABC = \angle BCA$:
$\angle ACD + \angle BCA + (\angle BCA + \angle ACD) = 180^\circ$
$2(\angle ACD + \angle BCA) = 180^\circ$
Step 5: Final Deduction
By observing the geometry, $(\angle ACD + \angle BCA) = \angle BCD$.
$2(\angle BCD) = 180^\circ$
Dividing both sides by $2$:
$\angle BCD = 90^\circ$
Final Solution: $\angle BCD = 90^\circ$, thus proving that $\triangle BCD$ is a right-angled triangle.
More Questions from Class 9 Mathematics Triangles EXERCISE 7.2
- Q1(i): In an isosceles triangle ABC, with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that : (i) $OB = OC$
- Q1(ii): In an isosceles triangle ABC, with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that : (ii) $AO$ bisects $\angle A$
- Q2: In $\triangle ABC$, $AD$ is the perpendicular bisector of $BC$ (see Fig. 7.30). Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$.
- Q3: ABC is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively (see Fig. 7.31). Show that these altitudes are equal.
- Q4(i): ABC is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig. 7.32). Show that (i) $\triangle ABE \cong \triangle ACF$
- Q4(ii): ABC is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig. 7.32). Show that (ii) $AB = AC$, i.e., ABC is an isosceles triangle.
- Q5: ABC and DBC are two isosceles triangles on the same base $BC$ (see Fig. 7.33). Show that $\angle ABD = \angle ACD$.
- Q7: ABC is a right angled triangle in which $\angle A = 90^{\circ}$ and $AB = AC$. Find $\angle B$ and $\angle C$.
- Q8: Show that the angles of an equilateral triangle are $60^{\circ}$ each.
CBSE Solutions for Class 9 Mathematics Triangles
Chapters in CBSE - Class 9 Mathematics
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