Question

Theorem:If two non-parallel sides of
a trapezium are equal, it is cyclic.

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Answer 1

Answer:

Heya friend,!!

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 ΔALD and ΔBMC,

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)

Hope it helps u

Step-by-step explanation:

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Answer 2

AnsweR :

$\bf{\Large{\underline{\sf{Given\::}}}}$

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

$\bf{\Large{\underline{\sf{To\:prove\::}}}}$

ABCD is a cyclic quadrilateral.

$\bf{\Large{\underline{\sf{Construction\::}}}}$

We draw DE ⊥ AB and CF ⊥ AB.

$\bf{\Large{\underline{\rm{\red{Proof\::}}}}}$

In ΔDEA and ΔCFB

AD = BC     [given]

∠DEA = ∠CFB        [each 90°]

BF = CF   [Altitude]

by R.H.S rule;

$\sf{\triangle DEA\cong \triangle CFB}$

_________________________

$\sf{\angle A\:=\:\angle B....................(1)}}\\\\\\\sf{\angle ADE\:=\:\angle FCB}\\\\\\\sf{\angle ADE+90\\\degree\:=\:\angleFCB+90\degree}\\\\\\\sf{\angle ACD\:=\:\angle BCD\implies\angle A\:=\:\angle B.........................(2)}$

Now,

Total angle of cyclic  = 360°

$\mapsto\sf{\angle A+\angle B+\angle C+\angle D=360\degree}\\\\\\\mapsto\sf{\angle B+\angle B+\angle D+\angle D=360\degree}\\\\\\\mapsto\sf{2\angle B+2\angle D=360\degree}\\\\\\\mapsto\sf{2(\angle B+\angle D)=360\degree}\\\\\\\mapsto\sf{\angle B+\angle D\:=\:\cancel{\dfrac{360\degree}{2} }}\\\\\\\mapsto\sf{\orange{\angle B+\angle D\:=\:180\degree}}$

We call it sum of one pair of opposite angles is 180°.

Thus,

$\bigstar$ABCD is a cyclic trapezium.