In a triangle ABC, D and E are points on BC such that AD - AF and angle BAD = angle CAt AB = (2p * 3) BD = 2p AC = (3q - 1) and C*Gamma = 4 then find the value of (p + q)
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We have a triangle ABC with point D on side BC. We're given that AB equals 6p, AC equals 3q minus 1, and BD equals 2p. The line AD is an angle bisector, meaning angle BAD equals angle CAD. We also have the condition C times gamma equals 4, and we need to find p plus q.
Since AD is an angle bisector, we can apply the Angle Bisector Theorem. This states that the ratio of the two sides of a triangle is equal to the ratio of the segments they create on the opposite side. So AB over AC equals BD over CD. Substituting our values: 6p over 3q minus 1 equals 2p over CD. Solving for CD, we get CD equals 3q minus 1 over 3.
Now we need to interpret the condition C times gamma equals 4. In the context of triangle geometry, this most likely means AC times BD equals 4. Substituting our known values: 3q minus 1 times 2p equals 4. Simplifying this equation: 2p times 3q minus 1 equals 4, which gives us p times 3q minus 1 equals 2.
Now we need to solve the equation p times 3q minus 1 equals 2. Looking for simple integer solutions, let's try p equals 1 and q equals 1. Checking: 1 times 3 times 1 minus 1 equals 1 times 2 equals 2, which works perfectly. Therefore, p plus q equals 1 plus 1 equals 2.
To summarize our solution: We applied the Angle Bisector Theorem to establish the relationship between the sides. We interpreted the condition C gamma equals 4 as AC times BD equals 4. This led us to the equation p times 3q minus 1 equals 2. Finding the simple solution p equals 1 and q equals 1, we conclude that p plus q equals 2.