Question

In a regular
the number of
and prove
pentagon ABCDE a calculate
degrees in the angle ABC...
PROVE
that BC||AD.​

Answer 1

Answer:

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Answer 2

Answer:

ABCDE is given to be regular pentagon.

We have n = 5

∴ Each interior angle = [(2n-4)/n]*90° = [(2*5 - 4)/5]*90° = 108°

∴ ∠ABC = 108°

Let’s draw a line from A to D.

Now, consider ∆AED, we have

AE = ED [∵ all sides of a regular pentagon are of equal length]

∴ ∆AED is isosceles triangle [∵ two sides of a triangle AED are equal]

∴ ∠EAD = ∠EDA

Also,  

∠EAD + ∠EDA + ∠AED = 180°

Or, ∠EAD + ∠EDA = 180° - 108° = 72°

∴ ∠EAD = ∠EDA = 72° / 2 = 36°

∴ ∠BAD = ∠EAB – ∠EAD = 108° - 36° = 72°

Similarly, ∠CDA = ∠EDC – ∠EDA = 108° - 36° = 72°

We know that, if a transversal intersects two lines so that the interior angles on the same side of the transversal are supplementary, then the two lines are parallel to each other i.e.,  

angle ABC + Angle BAD = 108° + 72° = 180°

angle BCD + Angle CDA = 108° + 72° = 180°

∴ BC // AD

Hence proved