Question

$ABC$ and $DBC$ are two isosceles triangles on the same side of $BC.$ Prove that :
$\left(i\right)DA\left(orAD\right)$ produced bisects $BC$ at right angle.
$\left(ii\right)\angle BDA=\angle CDA$

Answer 1

i) In a isosceles triangle , The perpendicular bisector Of the $3^{rd}$ side passes through Vertex

So, the line Passes through Both $A,D$ i.e., $AD$ when produced bisects $BC$ at right angle

ii) In $△ABD△ACD$

$AD=AD$ (common side)$\cdots \left(1\right)$

$DB=CD$ ($△BCD$ is isosceles )$\cdots \left(2\right)$

$AB=AC$ ($△ABC$ is isosceles )$\cdots \left(3\right)$

From $\left(1\right)\left(2\right)\left(3\right)⟹$ Triangles are congruent

So, $\angle BDA=\angle CDA$