Quadrilaterals — Question 6

(i) In Δ ABC,

It is given that M is the mid-point of AB and MD || BC.

By converse of mid-point theorem, we get :

Thus, D is the mid-point of AC.

Hence, proved that D is the mid-point of AC.

(ii) As DM || CB and AC is a transversal,

We know that,

Sum of co-interior angles = 180°.

⇒ ∠MDC + ∠DCB = 180°

⇒ ∠MDC + 90° = 180°

⇒ ∠MDC = 180° - 90°

⇒ ∠MDC = 90°

Hence, proved that MD ⊥ AC.

(iii) Join MC,

In Δ AMD and Δ CMD,

⇒ AD = CD (D is the mid-point of side AC)

⇒ ∠ADM = ∠CDM (Since, MD ⊥ AC)

⇒ DM = DM (Common side)

∴ Δ AMD ≅ Δ CMD (By S.A.S. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AM = CM (By C.P.C.T.)

We know that,

M is the mid-point of AB.

∴ AM = $\dfrac{1}{2}AB$

⇒ CM = AM = $\dfrac{1}{2}AB$.

Hence, proved that CM = MA = $\dfrac{1}{2}AB$.