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  • ABC is an isosceles triangle with AB equal to AC and D is a point on BC such that AD perpendicular BC (Figure). To prove that angle BAD equal to angle CAD, a student proceeded as follows: In triangle ABD and triangle ACD, AB equal to AC (Given) angle B

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD \perp BC (Figure). To prove that \angleBAD = \angleCAD, a student proceeded as follows:

In \triangleABD and \triangleACD,

AB = AC (Given)

\angleB = \angleC (because AB = AC)

and \angleADB = \angleADC

Therefore, \triangleABD \cong \triangleACD (AAS)

So, \angleBAD =\angleCAD (CPCT)

What is the defect in the above arguments?

Answers (1)

In \triangleABC, AB = AC (Given)

\angleACB = \angleABC          (angles opposite to equal side are equal)

In \triangleABD and \triangleACD

\angleABD = \angleACD         (given)

\angleABD = \angleACD         (from above)

\angleADB = \angleADC         (each 90°)

\therefore \triangleABD \cong \triangleACD

\angleBAD = \angleCAD

So, the defect in the above-given argument, first prove

\angleABD = \angleACD

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