Question

${\underline{\sf{\purple{\mathfrak{Question –}}}}}$

If the non- parallel sides of trapezium are equal, prove that it is cyclic?

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Answer 1

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$\sf\underbrace{ Question }$

• If the non- parallel sides of trapezium are equal, prove that it is cyclic?

$\bf\underline{Given -}$

• ABCD is a trapezium where AB || DC & non parallel sides are equal, i.e., AD = BC

$\bf\underline{To\:prove -}$

• ABCD is cyclic quadrilateral

$\bf\underline{Construction :}$

• We draw DE ⊥ AB & CF ⊥ AB

$\bf\underline{proof -}$

To prove ABCD is a cyclic quadrilateral, we prove that sum of one pair of opposite angles is 180°

• $⟹$In ∆ADE & ∆BCF

• $⟹$$\sf{∠AED\: =\: ∠BFC}$

• $⟹$$\sf{AD\: = \:BC}$

• $⟹$$\sf{DE\:= \:CF}$

• $⟹$$\sf{∴\: ∆ADE \:≈\: ∆BCF}$

• $⟹$$\sf{So,\: ∠DAE \:= \:∠CBF}$

• $⟹$$\sf{i.e,\:∠A\:=\:∠B}$

$\bf\underline{Now,}$

for parallel lines AB and DC & AD is the transversal line

• $⟹$$\sf{∠A\: + \:∠D \:=\: 180°}$

• $⟹$$\sf{∠B\:+\: ∠D \:= \:180°}$

So,

• In ABCD, sum of one pair of opposite angles is 180°

$\bf\underline{Therefore,}$

ㅤ$\large{ \underline{ \boxed{ \textsf{ABCD\:is \:a \:cyclic\:Quadrilateral }}}}$

$\bf\underline\blue{Hence\: proved}$${\boxed{\red{\checkmark{}}}}$

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Answer 2

Answer:

Question

• If the non- parallel sides of trapezium are equal, prove that it is cyclic?

Answer

Given :-

• Let us consider a trapezium ABCD in which AB || DC & non parallel sides AD = BC.

To prove : −

• ABCD is cyclic trapezium

Concept Used :-

• In order to prove that ABCD is a cyclic quadrilateral, we have to prove that sum of the pair of opposite angles is supplementary i.e. 180°

Construction :-

• Construct DE ⊥ AB & CF ⊥ AB

Proof :-

In ∆ADE & ∆BCF

⇛∠AED=∠BFC (each 90°)

⇛AD = BC (given)

⇛DE=CF (Distance between || lines)

∴∆ADE ≅ ∆BCF [RHS Congruency]

So,∠DAE =∠CBF (CPCT)

⟹ ∠DAB=∠CBA

Now, AB || CD & AD is the transversal line.

⟹∠DAB + ∠CDA = 180°

⟹∠CBA + ∠CDA = 180° ( as ∠DAB =∠CBA)

Therefore, ABCD is a cyclic trapezium.

$\bf\underline\red{Hence\: proved}$