In the given fig. ABCD is a parallelogram. A circle through A, B and C intersects CD produced at E. Prove that AD = AE.
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Meww:
Ur figure is wrong i think
no-no it's as per the question
yes correct figure
is my answer correct??
I hv seen this type of ques in which d was outside the figure and e was on the circumference
you have written ABED as a cyclic quadrilateral
jzt correct it as ABCE
THANQ
WELCOME
Answers
Answered by
1
ABCE IS A CYCLIC QUADRILATERAL
and ABCD is a parallelogram
As AB║CD or AB║CE
ABCE can be called a trapezium
and a cyclic trapezium is an isosceles trapezium
∴AE = BC___________(1)
and as AD =BC
∴AD = AE {FROM ()}
and ABCD is a parallelogram
As AB║CD or AB║CE
ABCE can be called a trapezium
and a cyclic trapezium is an isosceles trapezium
∴AE = BC___________(1)
and as AD =BC
∴AD = AE {FROM ()}
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Answered by
2
given : DC ║ AB
so, EC ║ AB
also, AD = BC
and ∠B + ∠C = 180°
but, ABDE is a cyclic quadrilateral so,
∠B + ∠E = 180°
⇒ ∠E = ∠C
Now, draw AL ⊥ EC & BM ⊥ EC
In tri. ALE & BMC,
∠E = ∠C
AL = BM (since the distance between ║ lines is equal)
∠ALE = ∠BMC (90°)
so by AAS criterion both are congruent.
⇒ AD = AE ( by C.P.C.T.)
so, EC ║ AB
also, AD = BC
and ∠B + ∠C = 180°
but, ABDE is a cyclic quadrilateral so,
∠B + ∠E = 180°
⇒ ∠E = ∠C
Now, draw AL ⊥ EC & BM ⊥ EC
In tri. ALE & BMC,
∠E = ∠C
AL = BM (since the distance between ║ lines is equal)
∠ALE = ∠BMC (90°)
so by AAS criterion both are congruent.
⇒ AD = AE ( by C.P.C.T.)
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It is not given that ABCE is a cyclic quadrilateral.
but we can see tht in the figure
but their points are concylic therefore ABCE is cyclic
Thanks !!
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