ABCD is a parallelogram. Point P divides AB in the ratio 2:3 and

Q) ABCD is a parallelogram. Point P divides AB in the ratio 2:3 and point Q divides DC in the ratio 4:1. Prove that OC is half of OA.

Ans:

Given that ABCD is a parallelogram.

Therefore, AB ǁ CD and BC ǁ AD

Step 1: Since, Point P divides AB in the ratio 2:3

Therefore, if AB = a, then:

AP = $\frac{2}{5}$ a

and BP = $\frac{3}{5}$ a

Step 2: Similarly, Point Q divides CD in the ratio 4:1

Therefore, since CD = AB = a, then:

DQ = $\frac{4}{5}$ a

and QC = $\frac{1}{5}$ a

Step 3: Let’s look at Δ AOP and Δ QOC,

∠ AOP = ∠ QOC (vertically opposite angles)

∠ OAP = ∠ QCO (interior angles)

By AA similarity rule,

Δ AOP $\sim$ Δ QOC

∴ $\frac{OA}{OC}$ = $\frac{AP}{QC}$

Step 4: Let’s start substituting values of AP and QC from previous steps:

We calculated: AP = $\frac{2}{5}$ a and QC = $\frac{1}{5}$ a

$\frac{OA}{OC}$ = $\frac{\frac{2}{5}a}{\frac{1}{5}a}$

$\frac{OA}{OC}$ = $\frac{2}{1}$

OC = $\frac{1}{2}$ OA

Therefore, it is proved that OC is half of OA.

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