In the given figure, ABC and 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) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:
More Solutions:
- AC is its diagonal. Show that
- Show that the quadrilateral formed.
- Prove that CF = 14 AC.
- The straight lines ED and EC.
- Prove that the line segment AD.
- AB || DC, E and F are mid-points of AD and BD.
- Prove that PQRS is a rhombus.
- G is mid-point of CD. Calculate:
- In a ∆ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm.
- If P and Q are mid-points of the sides BC and CD.