Question
✓
Answer
Given,
ABC and DBC are two isosceles triangles. To show,
$\angle\text{ABD}= \angle\text{ACD}$
In $\triangle\text{ABD}$ and $\triangle\text{ACD}$
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by SSS congruence condition.
Thus, $\angle\text{ABD}= \angle\text{ACD}$ (by CPCT).
ABC and DBC are two isosceles triangles. To show,
$\angle\text{ABD}= \angle\text{ACD}$
In $\triangle\text{ABD}$ and $\triangle\text{ACD}$
AD = AD (Common)
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by SSS congruence condition.
Thus, $\angle\text{ABD}= \angle\text{ACD}$ (by CPCT).
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