In the given figure, ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC and CD || BA. Show that
(i) ∠DAC=∠BCA
(ii) ABCD is a parallelogram.
Solution:
More Solutions:
- Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD.
- Prove that area of quad.
- Prove that, area of ∆ CPD = area of ∆ AQD.
- D, E and F are mid-points
- Prove that BCEF is a trapezium.
- Prove that two triangles having equal areas.
- Prove that area of ∆ ABD: area of ∆ ADC = m : n.
- Calculate the height of parallelogram.
- Prove that : area of ∆APQ = area of ∆DPQ = 16
- Prove that: area of || gm ABCD + area of || gm AEFB = area of || gm EFCD.