Question

$ABC$ is a triangle right angled at $C$. A line through the mid-point $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ and $D$. Show that
(i) $D$ is the mid-point of $AC$
(ii) $MB\perp AC$
(iii) $CM=MA=\frac{1}{2}AB$

Answer 1

(i) In $△ABC$,

$M$ is the mid point of $AB$ and $MD\parallel BC$.

Therefore, $D$ is the mid- point of $AC$.
i.e., $AD=DC$ ... (1)

(ii) Since $MD\parallel BC$, corresponding angles has to be equal

$\angle ADM=\angle ACB$

$\angle ADM={90}^{\circ }$

As $\angle ACB={90}^{\circ }$ is given

Since $\angle ADM$ and $\angle CDM$ are angles of a linear pair

But $\angle ADM+\angle CDM={180}^{\circ }$

Therfore ${90}^{\circ }+\angle CDM={180}^{\circ }\Rightarrow \angle CDM={90}^{\circ }$

Thus, $\angle ADM=\angle CMD={90}^o$ ... (2)

$\Rightarrow MD\perp AC$

(iii) In triangles $AMD$ and $CMD$, we have
$AD=CD$ (from(1))

$\angle ADM=\angle CDM$

and $MD=MD$ (Common)

Using SAS criterion of congruence

$△AMD\cong △CMD$

$\Rightarrow MA=MC$ (Since corresponding parts of congruent triangles are equal)
Also,

$MA=\frac{1}{2}AB$, since $M$ is the mid-point of $AB$.

Hence, $CM=MA=\frac{1}{2}AB$