Question

In this figure, AB and CD are parallel. M is the mid point of BC. Prove that the area of

▵DCM and ▵BMN are equal?​

Answer 1

Answer:

Given, lines AB and CD are parallel and M is the midpoint of AB.

As AB and CD are parallel, AD, DM, MC and CB are transversal.

(i) ∠AMD = ∠MBC = 60⁰ (corresponding angles)

∠CMB = ∠DAM = 40⁰ (corresponding angles)

∠CDM = AMD = 60⁰ (alternate interior angles)

∠DCM = ∠CMB = 40⁰ (alternate interior angles)

On straight line AMB,

∠ AMD + ∠DMC + ∠CMB = 180⁰ (angles in a straight line)

60⁰ + ∠DMC + 40⁰ = 180⁰

∠DMC + 100⁰ = 180⁰

∠DMC = 180⁰-100⁰

∠DMC = 80⁰

∠ADM = ∠DMC = 80⁰ (alternate interior angles)

∠MCB = ∠DMC = 80⁰ (alternate interior angles)

Therefore, now we have,

(see the image)

(ii) Quadrilaterals AMCD and MBCD both contain two equal triangles. That is what makes special.

Quad. AMCD consists of ∆AMD and ∆DMC.

Quad. MBCD consists of ∆MBC and ∆DMC.

Step-by-step explanation:

pls mark me as brainliest. Hope it helps you!