Question

Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.If AB = CD, then show that:
(i) ar ($△$DOC) = ar ($△$AOB) (ii) ar ($△$DCB) = ar ($△$ACB) [4 MARKS]

Answer 1

Statement of the theorem : 1 Mark
Application of the theorem : 1Mark
Calculation : 2 Marks

Let us draw DN $\perp$ AC and BM $\perp$ AC.

i.) In $△$NOD and $△$MOB,

$\angle$DNO = $\angle$BMO (By construction)

$\angle$DON = $\angle$BOM (Vertically opposite angles)

OD = OB (Given)

$△$DON $\cong$ $△$BOM [ AAS congruence rule]

$\therefore$ DN = BM ---------- (1)

We know that congruent triangles have equal areas.

$\therefore$ ar ($△$DON) = ar ($△$BOM) ------------(2)

In $△$DNC and $△$BMA,

$\angle$DNC = $\angle$BMA (By construction)

CD = AB (Given)

DN = BM [Using equation (1)]

So, $△$DNC $\cong$ $△$BMA [RHS congruence rule]

ar ($△$DNC) = ar ($△$BMA) --------- (3)

Adding (2) and (3)

ar ($△$DOC) = ar ($△$AOB)

ii.) We obtained,

ar ($△$DOC) = ar ($△$AOB)

Add ar ($△$OCB) on both sides

ar($△$DCB) = ar($△$ACB)