Line connecting the half of a triangle height with its side.

Question

The isosceles triangle ABC (AC=BC) has a height CC1. Point E is part of BC where EB = 2CE.
If AE crosses CC1 in point D, prove that CD=DC1.

Answer 1

I will use Mass Point Geometry.

C has mass 2 and B has mass 1 using the given ration.

AC1=BC1 as C1 is the foot of the altitude in an isosceles triangle where AC=BC, so CC1 is also a median.

Then, the mass of B is equal to the mass of A which must now be 1.

The mass of C1=massA+massB=2.

C has mass 2 and Cq has mass 2, equal masses mean CD=DC1.

You could also show the result using coordinate geometry, perhaps with origin at C1.

Answer 2

Let [.] denote triangle areas.

$$\frac{CD}{C_1D} = \frac{[ACE]}{[AC_1E]} = \frac{\frac13[ABC]}{\frac12[ABE]} =\frac{\frac13[ABC]}{\frac12\cdot\frac23[ABC]}=1$$

where we used the facts that triangles $ACE$ and $AC_1E$ share their base side $AE$.