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piyoosh roy

· started a discussion

· 1 Months ago

BCD=CAB ,,, HOW ????????? (Question: 107619.0 )

BCD=CAB ,,, HOW ?????????

Question:
In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then: ∠CAD + ∠CBD = ?

Options:
A) 120°
B) 90°
C) 360°
D) 180°
Solution:
Ans: (d)


\(\angle \)BCD = \(\angle \)CAB (i)---(Alternate segment theorem)

\(\angle \)BDC = \(\angle \)DAB (ii)---(Alternate segment theorem)

By adding (i) and (ii) we get:

\(\angle \)BCD +\(\angle \)BDC = \(\angle \)CAB + \(\angle \)BAD

∠BCD +∠BDC = ∠CAD ....(iii)

In \(\triangle\)BCD,

\(\angle \)BCD +\(\angle \)BDC + \(\angle \)CBD =1800   (Angle sum property of triangle)

\(\angle \)CAD + \(\angle \)CBD = 180° ---from (iii)

Knowledge Expert

· commented

· 1 Months ago

Dear student
BCD = CAB
By alternate segment theorem.
for more explanation please refer solution.

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