Question

In right $△ABC,\angle C={90}^0,M$ is the mid-point of hypotenuse A-B. C is jointed to M and produced to D such that DM= CM. Point D is jointed to B. Show that $△DBC\cong △ACB.$

Answer 1

$In△BDM\&△AMC$

$BM=CM\left(given\right)$

$\angle DMB=\angle AMC\left(verticallyopposite\right)$

$BM=AM\left(given\right)$

$\therefore △BDM\cong △AMCbyS.A.Scongruency$

$\therefore AC=DB(C.P.C.T)-\left(1\right)$

$and\angle ACM=\angle BDM(C.P.C.T)$

$and\angle CAM=\angle DBM(C.P.C.T)$

$\therefore AC||BD(\because alternateanglesequal)$

$\therefore \angle ACB=\angle DBC\left(correspondingangles\right)$

$(\because AC||BDandBCisthecommontransverse)$

$\therefore \angle DBC={90}^{\circ }$

$In△DBC\&△ACB$

$BC=CB\left(common\right)$

$\angle DBC=\angle ACB={90}^{\circ }\left(proved\right)$

$DB=AC\left(from1\right)$

$\therefore DBC\cong △ACB$

$Hence,proved$