State and prove the mid-point theorem.
Solution:
Statement:
The line joining the mid-points of two sides of a triangle is parallel to the third side and is equal to half of it.
Proof:
In ΔABC, E and F are mid-points of side AB and AC respectively.
Construction:
Draw CG||BA and produce EF to D such that D lies on CG.
In ΔAEF and ΔCDF,
∠EAF = ∠FCD (Alternate interior angles)
AF = FC (Since F is the mid-point of AC)
∠AFE = ∠DFC (Vertically opposite angles)
∴ ΔAEF ≅ ΔCDF (By ASA congruency criterion)
Now, we know that corresponding parts of congruent triangles are the same.
Therefore,
EF = DF and AE = DC
However, AE = BE
∴ BE = DC
Also, AB||CD gives BE||CD.
∴ In quadrilateral BCDE, one pair of opposite sides is equal and parallel.
∴ BCDE is a parallelogram.
∴ EF||BC
Now, ED = EF + FD
( EF = FD)
⇒ 2EF = ED
( ED =BC, as BCDE is a parallelogram)
Hence, proved
