ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that

(i) D is the midpoint of AC

(ii) MD ⊥ AC

(iii) CM = MA = AB

Given:- ACB is right angle triangle

M is midpoint of AB

DM||CB

Formula used:- Line drawn through midpoint of one side of triangle

Parallel to other side , bisect the 3^rd side

* SAS congruency property

If 2 sides and angle between the two sides of both

the triangle are equal, then both triangle are congruent

Solution:-

(1) In Δ ABC

M is midpoint of AB

And DM||CB

∴ D is midpoint of AC

AD=DC

∵ Line drawn through midpoint of one side of triangle

Parallel to other side , bisect the 3^rd side

(2) In Δ ABC

DM||CB

∠ ADM=∠ ACB [Corresponding angles]

∠ ACB =90°

∠ ADM=90°

(3) In Δ ADM and Δ DMC

⇒ DM=DM [Common in both triangles ]

⇒ AD=DC [D is the midpoint]

As ADC is straight line

∠ ADM+∠ MDC=180°

∠ MDC=180° - ∠ ADM

∠ MDC=90°

⇒ ∠ ADM=∠ MDC

Hence;

In Δ ADM ≅ Δ DMC

∴ CM=MA

⇒ MA=AB [M is midpoint of AB]

∴ CM=MA=AB