In a triangle ABC, D and E are points on BC such that AD = AE and ∠BAD = ∠CAE. IF AB = (2p +3), BD = 2p, AC = (3q -1) and CE = q, then find the value of (p+q).

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In \(\Delta \)ABD and \(\Delta \)AEC,

AD = AE [Given]

\(\angle\)ADB = \(\angle\)AEC  [Exterior angles of ADE, where \(\angle\)ADE = \(\angle\)AED]

\(\angle\)BAD = \(\angle\)CAE [Given]

So, \(\Delta \)ABD \(similar\) \(\Delta \)AEC

Then,

= \(\frac{AB}{AC}\) = \(\frac{BD}{CE}\) = \(\frac{AD}{AE}\)

= \(\frac{2p\;+\;3}{3q\;-\;1}\) = \(\frac{2q}{q}\) = 1 [as AD = AE]

By taking \(\frac{2q}{q}\) = 1

= 2p = q    ..(1)

By taking \(\frac{2p\;+\;3}{3q\;-\;1}\) = 1

= 2p + 3 = 3q - 1

= 3q - 2p = 3 + 1

= 3q - q = 4  [as 2p = q]

= 2q = 4

= q = \(\frac{4}{2}\) = 2

Put q = 2 in equation (1),

2p = 2

= p = 1

Then, (p + q) =  1 + 2 = 3

Therefore, (p+q) = 3