Question

Question 8 (iv)
In right triangle ABC, right angled 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 (see the given figure). Show that:
$CM=\frac{1}{2}AB$

Answer 1

$In\Delta AMCand\Delta BMDAM=BM\left(MismidpointofAB\right)\angle AMC=\angle BMD\left(verticallyoppositeangles\right)CM=DM\left(given\right)\therefore \Delta AMC\cong \Delta BMD\left(bySAScongruencerule\right)\therefore AC=BD\left(byCPCT\right)And\angle ACM=\angle BDM\left(byCPCT\right)Wehave\angle ACM=\angle BDMButACMandBDMarealternateinterioranglesSincealternateanglesareequal.Hence,wecansaythatDB||AC\Rightarrow \angle DBC+\angle ACB={180}^{\circ }(co-interiorangles)\Rightarrow \angle DBC+{90}^{\circ }={180}^{\circ }(\angle ACB={90}^{\circ })\Rightarrow \angle DBC={180}^{\circ }-{90}^{\circ }\Rightarrow \angle DBC={90}^{\circ }\Rightarrow DB=AC\left(ByCPCT\right)\dots \left(i\right)In\Delta DBCand\Delta ACBDB=AC\left(From\right(i\left)\right)BC=BC\left(Common\right)\angle DBC=\angle ACB={90}^{\circ }\therefore \Delta DBC\cong \Delta ACBbySAScongruence\Rightarrow CD=AB\left(ByCPCT\right)\Rightarrow 2CM=AB\Rightarrow CM=\frac{1}{2}AB$