Question 78
In $$\triangle ABC$$, AB = AC and D is a point on BC. If BD = 5 cm, AB = 12 cm and AD = 8 cm, then the length of CD is:
Solution
by stewart theorem,
$$(AB)^2CD + (AC)^2BD = BC[(AD)^2 + BD.CD]$$
$$(12)^2CD + (12)^2\times 5 = (5 + CD)[(8)^2 + 5CD]$$
$$144CD + 720 = (5 + CD)[64 + 5CD]$$
$$144CD + 720 = 320 + 25CD + 64CD + 5(CD)^2$$
$$5(CD)^2 -Â 55CD + 400 = 0$$
$$(CD)^2 - 11CD + 80 = 0$$
$$(CD)^2 - 16CD + 5CDÂ + 80 = 0$$
$$(CD - 16)(CD + 5) = 0$$
CD = 16 cm
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