The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Then, ΔABP≡ΔACP If the above statement is true then mention answer as 1, else mention 0 if false
Open in App
Solution
In △s, ABP and ACP, AP=AP (Common) AB=AC (Given, ABC is an isosceles triangle) ∠ABC=∠ACB (Isosceles triangle porperty) Thus, △ABP≅△ACP (SAS postulate)