Answer:
Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.
To prove: Point A,D,P are collinear.
Proof:
→Case 1.  When vertices A and P are opposite side of Base BC.
In Δ ABD and Δ ACD
AB= AC Â [Given]
AD is common.
BD=DC Â [median of a triangle divides the side in two equal parts]
Δ ABD ≅Δ ACD [SSS]
∠1=∠2 [CPCT].........................(1)
Similarly, Δ PBD ≅ Δ PCD [By SSS]
∠3 = ∠4 [CPCT].................(2)
But,  ∠1+∠2+ ∠3 + ∠4 =360° [At a point angle formed is 360°]
2 ∠2 + 2∠4=360° [using (1) and (2)]
∠2 + ∠4=180°
But ,∠2 and ∠4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.
So, point A, D, P lies on a line.
CASE 2.
When ΔABC and ΔPBC lie on same side of Base BC.
In ΔPBD and ΔPCD
PB=PC[given]
PD is common.
BD =DC [Median of a triangle divides the side in two equal parts]
ΔPBD ≅ ΔPCD  [SSS]
∠PDB=∠PDC [CPCT]
Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]
As PD and AD are medians to same base BC of ΔPBC and ΔABC.
∴ P,A,D lie on a line i.e they are collinear.
