Triangles — Question 8

(i) In Δ AMC and Δ BMD,

⇒ AM = BM (M is the mid - point of AB)

⇒ ∠AMC = ∠BMD (Vertically opposite angles are equal)

⇒ CM = DM (Given)

∴ Δ AMC ≅ Δ BMD (By S.A.S. congruence rule)

Hence, proved that Δ AMC ≅ Δ BMD.

(ii) Since,

Δ AMC ≅ Δ BMD

∴ ∠ACM = ∠BDM (By C.P.C.T.)

From figure,

∠ACM and ∠BDM are alternate interior angles. Since alternate angles are equal, it can be said that DB || AC.

We know that,

Sum of co-interior angles = 180°.

⇒ ∠DBC + ∠ACB = 180°

⇒ ∠DBC + 90° = 180° [Since, ΔACB is right angled triangle at point C]

⇒ ∠DBC = 180° - 90°

∴ ∠DBC = 90°.

Hence, proved that ∠DBC is a right angle.

(iii) Since,

Δ AMC ≅ Δ BMD

∴ DB = AC (By C.P.C.T.)

In Δ DBC and Δ ACB,

⇒ DB = AC (Proved above)

⇒ ∠DBC = ∠ACB (Both equal to 90°)

⇒ BC = CB (Common)

∴ Δ DBC ≅ Δ ACB (By S.A.S. congruence rule)

Hence, proved that Δ DBC ≅ Δ ACB.

(iv) Since,

Δ DBC ≅ Δ ACB

⇒ AB = DC (By C.P.C.T.)

⇒ $\dfrac{1}{2}AB$ = $\dfrac{1}{2}DC$

It is given that M is the midpoint of DC

⇒ CM = $\dfrac{1}{2}DC$ = $\dfrac{1}{2}AB$

∴ CM = $\dfrac{1}{2}AB$

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