In the adjoining figure, AB || CD. Prove that p+q = r.

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Solution

Since AB || CD
BC is a tranversal
Thus $∠ B = ∠ C$ [Alternate $∠ a$ ]
Similarly, $∠ A = ∠ D$ [Alternate $∠ a$ ]
Thus, $p + q + x = 180 \Rightarrow x = 180 - (p + q)$
Also, x + r =180
$180 - (p + q) + r = 180 \Rightarrow - (p + q) + r = 0$
$\Rightarrow r = p + q$

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In the adjoining figure, AB || CD. Prove that p+q = r.

Since AB || CD BC is a tranversal Thus ∠B=∠C [Alternate ∠a] Similarly, ∠A=∠D [Alternate ∠a] Thus, p+q+x=180⇒x=180−(p+q) Also, x + r =180 180−(p+q)+r=180⇒−(p+q)+r=0 ⇒r=p+q

Since AB || CD
BC is a tranversal
Thus $∠ B = ∠ C$ [Alternate $∠ a$ ]
Similarly, $∠ A = ∠ D$ [Alternate $∠ a$ ]
Thus, $p + q + x = 180 \Rightarrow x = 180 - (p + q)$
Also, x + r =180
$180 - (p + q) + r = 180 \Rightarrow - (p + q) + r = 0$
$\Rightarrow r = p + q$