In the given figure- - ABC - 65- - BCE - 30- - DCE - 35- and - CEF - 145- Prove that AB - EF-

∠ ABC =65^∘ nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ∠ BCD nbsp;= ∠ BCE+ ∠ ECD nbsp;=30^∘ + 35^∘ = 65^∘ nbsp; nbsp; n ...

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Byju's Answer

Standard IX

Mathematics

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In the given ...

Question

In the given figure, $∠ A B C = 65^{\circ}, ∠ B C E = 30^{\circ}, ∠ D C E = 35^{\circ}$ and $∠ C E F = 145^{\circ}$ . Prove that $A B ∥ E F$ .

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Solution

$∠ A B C = 65^{\circ}$

$∠ B C D = ∠ B C E + ∠ E C D = 30^{\circ} + 35^{\circ} = 65^{\circ}$
$∠ A B C = ∠ B C D$ (Altemate Angles)

$\to A B ∥ C D$ ...(1)

$∠ F E C + ∠ E C D = l 45^{\circ} + 35^{\circ} = 180^{\circ}$

But these angles are consecutive interior angles formed on the same side of the transversal.

$\to C D ∥ E F$ ...(2)

From (1) and (2),

$A B ∥ E F$

Hence, proved.

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In the given figure, ∠ABC=65∘,∠BCE=30∘,∠DCE=35∘ and ∠CEF=145∘. Prove that AB∥EF.

∠ABC=65∘ ∠BCD=∠BCE+∠ECD=30∘+35∘=65∘ ∠ABC=∠BCD (Altemate Angles) →AB∥CD ...(1) ∠FEC+∠ECD=l45∘+35∘=180∘ But these angles are consecutive interior angles formed on the same side of the transversal. →CD∥EF ...(2) From (1) and (2), AB∥EF Hence, proved.

In the given figure, $∠ A B C = 65^{\circ}, ∠ B C E = 30^{\circ}, ∠ D C E = 35^{\circ}$ and $∠ C E F = 145^{\circ}$ . Prove that $A B ∥ E F$ .