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Question 3
ABC is an isosceles triangle with AB = AC and D is a point on BC such that ADBC (see figure). To prove that BAD=CAD, a student proceeded as follows.

In ΔABD and ΔACD, we have AB= AC [given]
B=C [because AB=AC]
And ADB=ADC
Therefore,ΔABDΔACD [by AAS congruence rule]
So,BAD=CAD [by CPCT]
What is the defect in the above arguments?

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Solution

In ΔABC,AB=AC
ACB=ABC [angle oppsoite to the equal sides are equal]
In ΔABD and ΔACD,

AB=AC [given]
ABD=ACD [proved above]
ADB=ADC [each 90]
ΔABDΔACD [by AAS]
So, BAD=CAD [by CPCT]
So, the defect in the given argument is that firstly prove ABD=ACD
Hence, ABD=ACD is defect.

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