Question

In right triangle $ABC$, right angle is at $C,M$ is the mid-point of hypotenuse $AB,C$ is joined to $M$ and produced to a point $D$ such that $DM=CM$. Point $D$ is joined to point $B$. Show that:
$CM=\frac{1}{2}AB$

Answer 1

Given : Right angle $ABC$, right angle is at C,M is the mid-point of hypotenuse $AB$ and $DM=CM$ .

To show : $CM=AB/2$

Proof :

In triangle $DBM$ and triangle $AMC$ ,

$DM=MC$ (given)

$BM=MA$ (given)

$\angle DMB=\angle AMC$ (vertically opposite angle)

Therefore by SAS criterion of congruency , triangle $DMB$ is congruent to triangle $AMC$.

So,

$DB=AC$ (by CPCT)

$\angle BDM=\angle CAM$ (by CPCT)

Now, in triangle $DBC$ and $ABC$,

$DB=AC$ (proved above)

$BC=BC$ (common)

$\angle BDM=\angle CAM$ (proved above)

Therefore , by SSA criterion of congruency, triangle $DBC$ is congruent to triangle $ABC$ .

Therefore ,

$CD=AB$ (by CPCT)

$CM+MD=AB$ (because $CD=CM+MD$)

$CM+CM=AB$ (because M is midpoint of $CD$)

$2CM=AB$

$CM=AB/2$