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In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Answers (1)

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 =BD = \frac{1}{2} AC

Proof:

In \triangleADB and \triangleEDC

AD = CD                                 (D is midpoint)

BD = DE                                 (D is midpoint)

\angleADB = \angleEDC          (Vertically opposite angles)

\angleADB \cong \triangleEDC          (by SAS congruence)

\therefore AB = EC                             (by CPCT)

Now, \angleABC + \angleBCE = 180°

Þ 90° + \angleBCE = 180° (\angleABC = 90°, Given)

Þ\angleBCE = 90°

And EC || AB              (\because \angleBAD and \angleDCE are alternate angle)

In \triangleABC and \angleEBC

BC = BC                                 (common)

AB = EC                                 (From above)

\angleABC = \angleBCE          (From above)

\therefore \triangleABC \cong \triangleECB        (by ASA congruence)

AC = EB                                 (by CPCT)
\frac{1}{2}AC = \frac{1}{2}EB\\ \frac{1}{2}AC = BD

Hence proved

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