`triangle` ABC and `triangle` DBC are isosceles triangles on the same base BC. Prove that line AD bisects BC at right angles

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 (see the given figure). If AD is extended to intersect BC at P, show that, ( i ) triangle ABD = triangle ACD (ii) triangle ABP = triangle ACP (iii) AP bisects angle A as well as angle D. (iv) AP is the perpendicular bisector of BC

ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that angle ABD = angle ACD .

ABC is an isosceles triangle right angled at C. Prove that AB^(2) = 2AC^(2) .

In the given figure, ABC and DBC are two triangles on the same base BC. Prove that (ar(ABC))/(ar(DBC))= (AO)/(DO) .

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that: ( i ) AD bisects BC ( ii) AD bisects angle A

ABC is an isosceles triangle, with AC= BC. If AB^(2)= 2AC^(2) , prove that ABC is a right triangle.

ABC is an isosceles triangle, right angled at C. Prove that AB^(2)= 2AC^(2) .

In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ar(ABC))/(ar(DBC))= (AO)/(DO) .

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CBD.

In triangle ABC, AD is a median. Prove that AB + AC gt 2AD