$In the adjoining figure, □ ABCD is a square . △ BCE on side BC and △ ACF on the diagonal AC are similar to each other . S how that A (△ BCE) = \frac{1}{2} A (△ ACF) .$

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Solution

$Given : □ ABCD is a square . ∴ AC = \sqrt{2} BC . . . (1) [∵ Diagonal of a square = \sqrt{2} \times side of the square] △ BCE ~ △ ACF [Given] ∴ \frac{A (△ BCE)}{A (△ ACF)} = \frac{{(BC)}^{2}}{{(AC)}^{2}} . . . (2) \Rightarrow \frac{A (△ BCE)}{A (△ ACF)} = \frac{{BC}^{2}}{{(\sqrt{2} BC)}^{2}} [From (1) & (2)] \Rightarrow \frac{A (△ BCE)}{A (△ ACF)} = \frac{{BC}^{2}}{2 {BC}^{2}} \Rightarrow \frac{A (△ BCE)}{A (△ ACF)} = \frac{1}{2} ∴ A (△ BCE) = \frac{1}{2} A (△ ACF) Hence proved .$

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In the adjoining figure, □ABCD is a square. △BCE on side BC and △ACF on the diagonal AC are similar to each other. Show thatA(△BCE)=12A(△ACF).

Given: □ABCD is a square. ∴ AC=2 BC ...(1) [∵ Diagonal of a square =2 ×side of the square ]△BCE~△ACF [Given ]∴A(△BCE)A(△ACF)=BC2AC2 ...(2) ⇒A(△BCE)A(△ACF)=BC22BC2 [From (1) &(2)]⇒A(△BCE)A(△ACF)=BC22BC2⇒A(△BCE)A(△ACF)=12 ∴ A(△BCE)=12 A(△ACF) Hence proved.

$In the adjoining figure, □ ABCD is a square . △ BCE on side BC and △ ACF on the diagonal AC are similar to each other . S how that A (△ BCE) = \frac{1}{2} A (△ ACF) .$