In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Given : ABC is right angle triangle
D is mid point of AC
Construction: Construct EC parallel to AB, AE parallel to BC. Join DE.
To prove =
Proof:
In ADB and EDC
AD = CD (D is midpoint)
BD = DE (D is midpoint)
ADB = EDC (Vertically opposite angles)
ADB EDC (by SAS congruence)
AB = EC (by CPCT)
Now, ABC + BCE = 180°
Þ 90° + BCE = 180° (ABC = 90°, Given)
ÞBCE = 90°
And EC || AB ( BAD and DCE are alternate angle)
In ABC and EBC
BC = BC (common)
AB = EC (From above)
ABC = BCE (From above)
ABC ECB (by ASA congruence)
AC = EB (by CPCT)
Hence proved