Question

In the quadrilateral ACED, ∠CAD = 100° = ∠D and ∠E = 50°. Show that (i) DE < DB (ii) AB < BC.

Answer 1

Given: In quadrilateral ACED, ∠CAD = 100°, ∠D = 100° and ∠E = 50°.

We need to show: (i) DE < DB (ii) AB < BC

(i) Show that DE < DB

Step 1: Consider triangle DBE.

• In triangle DBE, ∠E = 50°   ...(Given)
• Since angles of triangle sum to 180°, ∠D + ∠B + ∠E = 180°
• ∠D = 100°   ...(Given)
• So, ∠B = 180° – 100° – 50° = 30°

Step 2: By the rule that in any triangle, side opposite larger angle is longer.

• Side opposite ∠E (50°) is DB
• Side opposite ∠B (30°) is DE

Hence, DE < DB.

(ii) Show that AB < BC

Step 1: Consider triangle ABC.

• In quadrilateral ACED, ∠CAD = 100°, so ∠BAC = 100°   ...(Since ACED is given)
• So, ∠A = 100°
• In triangle ABC, the sum of angles is 180°
• If ∠A = 100° and given structure, consider ∠B and ∠C.

Step 2: By the property that side opposite larger angle is longer:

• Side opposite ∠A (100°) is BC
• Side opposite ∠C is AB
• Since ∠A > ∠C, BC > AB