\(\triangle ABC\) and \(\triangle DBC\) are two isosceles triangles on the same base \(BC\) and vertices \(A\) and \(D\) are on the same side of \(BC\). If \(AD\) is extended to intersect \(BC\) at \(P\), show that (i) \(\triangle ABD \cong \triangle ACD\), (ii) \(\triangle ABP \cong \triangle ACP\), (iii) \(AP\) bisects \(\angle A\) as well as \(\angle D\), (iv) \(AP\) is the perpendicular bisector of \(BC\).
\(AD\) is an altitude of an isosceles triangle \(ABC\) in which \(AB = AC\). Show that (i) \(AD\) bisects \(BC\), (ii) \(AD\) bisects \(\angle A\).
Two sides \(AB\) and \(BC\) and median \(AM\) of one triangle \(ABC\) are respectively equal to sides \(PQ\) and \(QR\) and median \(PN\) of \(\triangle PQR\). Show that (i) \(\triangle ABM \cong \triangle PQN\), (ii) \(\triangle ABC \cong \triangle PQR\).
From (i), \(\angle ABM = \angle PQN\)
\(BE\) and \(CF\) are two equal altitudes of a triangle \(ABC\). Using RHS congruence rule, prove that the triangle \(ABC\) is isosceles.
\(ABC\) is an isosceles triangle with \(AB = AC\). Draw \(AP \perp BC\) to show that \(\angle B = \angle C\).