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ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, a student proceeded as follows:
In ∆ ABD and ∆ ACD,
AB = AC (Given)
∠B = ∠C (because AB = AC) and
∠ADB = ∠ADC
Therefore, ∆ ABD ≅ ∆ ACD (AAS)
So, ∠BAD = ∠CAD (CPCT)
What is the defect in the above arguments?[Hint: Recall how ∠B = ∠C is proved when AB = AC]

Solution:

Given, ABC is an isosceles triangle with AB = AC

D is a point on BC such that AD ⊥ BC

We have to prove that ∠BAD = ∠CAD

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, a student proceeded as follows: In ∆ ABD and ∆ ACD,AB = AC (Given) ,∠B = ∠C (because AB = AC) and ∠ADB = ∠ADC. Therefore, ∆ ABD ≅ ∆ ACD (AAS). So, ∠BAD = ∠CAD (CPCT). What is the defect in the above arguments.[Hint: Recall how ∠B = ∠C is proved when AB = AC]

In triangle ABC,

We know that the angles opposite to the equal sides are equal.

∠ACB = ∠ABC

Considering triangles ABD and ACD,

Given, AB = AC

Since AD ⊥ BC, ∠D = 90°

∠ADB = ∠ADC

AAS criterion (Angle-angle-side) states that when two angles and a non-included side of a triangle are equal to the corresponding angles and sides of another triangle, then the triangles are said to be congruent.

By AAS criterion, the triangles ABD and ACD are congruent

By CPCTC,

∠BAD = ∠CAD

Therefore, the defect in the given argument is ∠ABD = ∠ACD

✦ Try This: Determine the value of x in the given figure?

Determine the value of x in the given figure

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7

NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 3

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, a student proceeded as follows: In ∆ ABD and ∆ ACD,AB = AC (Given) ,∠B = ∠C (because AB = AC) and ∠ADB = ∠ADC. Therefore, ∆ ABD ≅ ∆ ACD (AAS). So, ∠BAD = ∠CAD (CPCT). What is the defect in the above arguments.[Hint: Recall how ∠B = ∠C is proved when AB = AC]

Summary:

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, the defect in the given argument is ∠ABD = ∠ACD

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