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:
$△DBC\cong △ACB$

Answer 1

Given:- BM$=$AM

DM$=$CM

Also, $\angle DMB=\angle CMA$(V.O.A)

Thus, $\Delta DMB\cong \Delta CMA$ by SAS criterion

Hence, by cpct,

$DB=AC$

& $\angle BDM=\angle ACM$

Thus $DB||AC$

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

Now, in $\Delta DBC$ & $\Delta ACB$

$DB=AC$ (proved above)

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

$BC=CB$ (common side)

$\therefore \Delta DBC\cong \Delta ACB$ by SAS criterion

Hence, proved.