Mathematics

On the sides, AB and AC of triangle ABC, equilateral triangles ABD and ACE are drawn. Prove that:  $$\angle {\text{CAD}} = \angle {\text{BAE}}$$


SOLUTION
Given:$$\triangle{ABD}$$ and $$\triangle{ACE}$$ are equilateral triangle.

$$\therefore$$ In $$\triangle{ABD}$$

$$\angle{BAD}=\angle{ABD}=\angle{BDA}={60}^{\circ}$$  .....$$(1)$$

In $$\triangle{ACE}$$

$$\angle{AEC}=\angle{ACE}=\angle{ACE}={60}^{\circ}$$  .....$$(2)$$

To prove:$$\angle{CAD}=\angle{BAE}$$

Proof:$$\angle{CAD}=\angle{BAC}+\angle{DAB}$$

$$\Rightarrow \angle{CAD}=\angle{BAC}+{60}^{\circ}$$ using $$(1)$$

$$\Rightarrow \angle{CAD}=\angle{BAC}+\angle{CAE}$$ using $$(2)$$

$$\Rightarrow \angle{CAD}=\angle{BAE}$$

Hence proved.
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