question 22
two circles ABCD and ABEF intersect at points A and B if CBA and DAF are straight lines prove that CD is parallel to EF
Answers
Answer:
Given: Two circles ABCD and ABEF intersecting at
A and B. DAF and CBE are straight lines.
To prove that: CD and EF are parallel.
Construction: Join ABCD to form a cyclic quadrilateral. Join ABEF to form another cyclic
quadrilateral.
Proof: In circle ABCD, DA is extended to F.
Now <FAB + <DAB = 180 being supplementary angles. Also ABCD being a cyclic quadrilateral, <DAB + <BCD = 180 deg. Hence <FAB = <BCD ... (1)
In circle ABEF, <FAB + <BEF = 180 ...(2) as ABEF is a cyclic quadrilateral.
From (1) and (2) we find <BCD + <B F = 180 deg, Since <BCD and <BEF are on the same side of the line CBE, CD must be parallel to EF. QED.
hope it helps...
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Answer:−
⏩ We have,
AB=AC and CE=BD
In ∆'s ABD and ACE, we have
→ AB=AC (Given)
→angle ABD= angle ACE (angles in the same segment)
→BD= CE (Given)
So, by SAS congruence criterion, we obtain
∆ ABD is congruent to ∆ ACE
=> AD= AE (CPCT)
✔✔✔✔✔✔✔✔
Hope it helps...:-)
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