In the given figure , DE is parallel to AC , express x in terms of a , b and y .
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We have ∠ BAC = ∠ BED because both are corresponding angles made by the transversal BE with the lines AC and ED and are also equal.
So, AC II ED. Since AC II ED and BD is a transversal. Then, ∠ BCA = ∠ BDE
corresponding angles in Δ BCA and Δ BDE.
∠ BAC = ∠ BED and given ∠ BCA = ∠ BDE
Proved above ⇒ Δ BCA is same as Δ BDE AA similarity.
⇒ BCBD = CADE = BABE Corresponding sides of similar triangles are proportional.
⇒ CADE = BABE
⇒ xa = yb + y
⇒ x = ayb + y
So, AC II ED. Since AC II ED and BD is a transversal. Then, ∠ BCA = ∠ BDE
corresponding angles in Δ BCA and Δ BDE.
∠ BAC = ∠ BED and given ∠ BCA = ∠ BDE
Proved above ⇒ Δ BCA is same as Δ BDE AA similarity.
⇒ BCBD = CADE = BABE Corresponding sides of similar triangles are proportional.
⇒ CADE = BABE
⇒ xa = yb + y
⇒ x = ayb + y
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