Question

In right triangle $ABC$, right angled at $C$, $M$ is the mid -point of hypotenuse $AB$. $C$ joined to $M$ and produced to a point $D$ such that $DM=CM$. Point $D$ is joined to point $B$ . Show that:
(i) $△AMC\cong BMD$
(ii) $\angle DBC$ is a right angle.
(iii) $△DBC\cong △ACB$
(iv) $CM=\frac{1}{2}AB$

Answer 1

In $△ABC,\angle C={90}^o$

$\to$ In $△DMB$ & $△AMC$

$DM=MC$ (Given)

$AM=MC$ (Mid point of $AB$ is $M$)

$\angle DMB=\angle AMC=\left(VOA\right)$ (Velocity opposite angle)

$\therefore$\triangle DMB\cong \triangle AMC$(Anu$SAS$ rule

then,

$DB=AC$ (By $CPCT$)

As $M$ is the mid point of $AB$, so $MC$ bisect $\angle C$ in the $△ABC$,

$\angle MCB=\angle MCA={45}^o$

$\angle MCA=\angle MOB$ (Alternate angles)

$\angle MDB={45}^o=\angle CDB$

& $\angle MCB=\angle DCB={45}^o$

in $△DBC$,

$\angle CDB+\angle DBC+\angle DCB={180}^o$ [sum of angle in $△=180$]

${45}^o+{45}^o+\angle DBC={180}^o$

$\angle DBC={90}^o$

$\therefore \angle DBC$ is a right angle

In $△DBC$ & $△ACB$,

$BC=BC$ (common )

$\angle ACB=\angle DBC={90}^o$

$DB=AC$ (proved through $CPCT$)

$\therefore △DBC\cong △ACB$ (By $SAS$ rule)

$AB=CD$ (By $CPCT$)

$MC=\frac{1}{2}\left(CD\right)$ (as $DM=MC$)

$=\frac{1}{2}\left(AB\right)$ (taken from put $\left(w\right)$)

Hence proved.