Question

$\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{D}$is a point on $\mathrm{BC}$ such that $\mathrm{AD}\perp \mathrm{BC}$ (Figure. To prove that $\angle \mathrm{BAD}=\angle \mathrm{CAD}$, a student proceeded as follows: In$\mathrm{\Delta }\mathrm{ABD}$ and $\mathrm{\Delta }\mathrm{ACD},\mathrm{AB}=\mathrm{AC}$ (Given) $\angle \mathrm{B}=\angle \mathrm{C}$ (because $\mathrm{AB}=\mathrm{AC}$) And $\angle \mathrm{ADB}=\angle \mathrm{ADC}$. Therefore, $\mathrm{\Delta }\mathrm{ABD}\approx \mathrm{\Delta }\mathrm{ACD}\left(\mathrm{AAS}\right)$So, $\angle \mathrm{BAD}=\angle \mathrm{CAD}$(CPCT). What is the defect in the above arguments?

[Hint: Recall how $\angle \mathrm{B}=\angle \mathrm{C}$ is proved when $\mathrm{AB}=\mathrm{AC}$].

Answer 1

Similar Triangles:

Two triangles are said to be similar if

(i) their corresponding angles are equal.

(ii) their corresponding sides are in proportion or the same ratio.

Proof given by the student for the isosceles triangle $\mathrm{ABD}$:

In$\mathrm{\Delta }\mathrm{ABD}$ and $\mathrm{\Delta }\mathrm{ACD},\mathrm{AB}=\mathrm{AC}$ (Given)

$\angle \mathrm{B}=\angle \mathrm{C}$ (because $\mathrm{AB}=\mathrm{AC}$)

And $\angle \mathrm{ADB}=\angle \mathrm{ADC}$.

Therefore, $\mathrm{\Delta }\mathrm{ABD}\approx \mathrm{\Delta }\mathrm{ACD}\left(\mathrm{AAS}\right)$So, $\angle \mathrm{BAD}=\angle \mathrm{CAD}$(CPCT).

Defect in the proof of the student:

Student stated that, $\angle \mathrm{B}=\angle \mathrm{C}$ (because $\mathrm{AB}=\mathrm{AC}$).

This statement is not true.

$\mathrm{AB}=\mathrm{AC}$ is not enough to conclude that, $\angle \mathrm{B}=\angle \mathrm{C}$.

Proving the similarity of $\mathrm{\Delta }\mathrm{ABD}$and $\mathrm{\Delta }\mathrm{ACD}$:

From the given details,

In$\mathrm{\Delta }\mathrm{ABD}$ and $\mathrm{\Delta }\mathrm{ACD}$

$\mathrm{D}$is a point on $\mathrm{BC}$ such that $\mathrm{AD}\perp \mathrm{BC}$

$\angle \mathrm{ADB}=\angle \mathrm{ADC}=90°$

$\mathrm{AB}=\mathrm{AC}$ (Given)

$\mathrm{AD}=\mathrm{AD}$ (Common Side)

RHS similarity condition:

Any two right-angled triangles are said to be similar if any one side and hypotenuse of one triangle is proportional to the one side and hypotenuse of the other triangle.

Thus, by RHS similarity criteria,$\mathrm{\Delta }\mathrm{ABD}$ is similar to $\mathrm{\Delta }\mathrm{ACD}$.

To prove $\angle \mathrm{BAD}=\angle \mathrm{CAD}$:

$\mathrm{\Delta }\mathrm{ABD}\approx$ $\mathrm{\Delta }\mathrm{ACD}$

If two triangles are said to be similar, then the corresponding parts of congruent triangles are equal.

Thus, $\angle \mathrm{BAD}=\angle \mathrm{CAD}$(By CPCT rule).

Hence, the defect in the given argument is $\angle \mathrm{B}=\angle \mathrm{C}$ (because $\mathrm{AB}=\mathrm{AC}$).