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Question 5In ...

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Question 5
In the following figure, DE||OQ and DF||OR, show that EF||QR.

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Solution

$I n Δ P Q O, D E | | O Q (G i v e n)$

$∴ \frac{P D}{D O} = \frac{P E}{E Q} . . . (i) [B y u s i n g B a s i c P r o p o r t i o n a l i t y T h e o r e m]$

In $Δ P O R, D F | | O R (G i v e n)$

$∴ \frac{P D}{D O} = \frac{P F}{F R} . . . (i i) [B y u s i n g B a s i c P r o p o r t i o n a l i t y T h e o r e m]$

From equation $(i)$ and $(i i)$ , we get,
$\frac{P E}{E Q} = \frac{P F}{F R}$

$S o, Δ P Q R, E F | | Q R [B y c o n v e r s e o f B a s i c P r o p o r t i o n a l i t y T h e o r e m]$

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Similar questions

Question 5
In the following figure, DE||OQ and DF||OR, show that EF||QR.

In the following figure, DE || OQ and DF || OR, show that EF || QR.

D E | | O Q

D F | | O R

E F | | Q R

.

In the figure, $D E ∥ O Q$ and $D F ∥ O R$ prove that $E F ∥ Q R$

Q. Question 6
If in two

Δ D E F a n d Δ P Q R, ∠ D = ∠ Q a n d ∠ R = ∠ E

, then which of the following is not true?

(A)

\frac{E F}{P R} = \frac{D F}{P Q}

\frac{D E}{P Q} = \frac{E F}{R P}

\frac{D E}{Q R} = \frac{D F}{P Q}

\frac{E F}{R P} = \frac{D E}{Q R}

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Converse of BPT

Standard X Mathematics

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