Question

Show that the line segment joining the mid-points of two sides of a triangle is half

of the third side. (Hint: Place triangle ABC in a clever way such that A is (0,0), B is

(2a,0) and C to be (2b,2c). Now consider the line segment joining the mid-points of

AC and BC. This will make calculations simple

Answer 1

Answer:

With only basic geometry:

If you've already studied similarity of triangles it is pretty easy: comparing triangles ΔCAB,ΔCPQ :

12=CPCA=CQCB,and the angle∠Cis common to both triangles

By similarity theorem , ΔCAB∼ΔCPQ , and thus

PQAB=12⟺2PQ=AB

That PQ||AB follows from the fact that similar triangles have the same angles, and thus ∠CAB=∠CPQ .

With vectors:

Put u:=CA→,v:=CB→ , then we get:

CP→=12u,CQ→=12CB,AB→=−u+v=−(v−u)

so

PQ→=−12+12b=−12(u−v)=12AB→

and we're done as the last line both proves the middle segment is parallel to AB and its length is half that of the latter.