If the non parallel sides of a trapezium are equal,prove that it is cyclic.

Asked by jauzijaved...(student), on 26/11/10

#### Answers

EXPERT ANSWER
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ΔALD and
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ΔBMC,

Hi!

Here is the answer to your question.

**Given**: ABCD is a trapezium where AB||CD and AD = BC

**To prove**: ABCD is cyclic.

**Construction**: Draw DL⊥AB and CM⊥AB.

**Proof**: In

AD = BC (given)

DL = CM (distance between parallel sides)

∠ALD = ∠BMC (90°)

ΔALD ≅ ΔBMC (RHS congruence criterion)

⇒ ∠DAL = ∠CBM (C.P.C.T) (1)

Since AB||CD,

∠DAL + ∠ADC = 180° (sum of adjacent interior angles is supplementary)

⇒ ∠CBM + ∠ADC = 180° (from (1))

⇒ ABCD is a cyclic trapezium (Sum of opposite angles is supplementary)

Cheers!

Posted by Sunanta Meitei(MeritNation Expert), on 27/11/10

This conversation is already closed by Expert

ith AB | |CD and BC = AD.

Draw AM ⊥ CD and BN ⊥ CD.

In ΔAMD and ΔBNC,

AD = BC (Given)

∠AMD = ∠BNC (By construction, each is 90°)

AM = BM (Perpendicular distance between two parallel lines is same)

∴ ΔAMD ≅ ΔBNC (RHS congruence rule)

∴ ∠ADC = ∠BCD (CPCT) ... (1)

∠BAD and ∠ADC are on the same side of transversal AD.

∠BAD + ∠ADC = 180° ... (2)

∠BAD + ∠BCD = 180° [Using equation (1)]

This equation shows that the opposite angles are supplementary.

Therefore, ABCD is a cyclic quadrilateral.

Posted by rubina.knisha.....(student), on 2/3/11

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