Two isosceles triangles share the same base. prove that the medians to this base are collinear. (there are two cases in this problem: vertices are on the same side of the base and vertices are on the opposite sides of the base! )

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   [Given]

AD is common.

BD=DC  [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.

c

step-by-step explanation:

at least i think its

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl