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Question 10
ABC is a triangle with BC=5. D is the foot of the perpendicular from A on BC. E is a point on CD such that BE=3. The value of $$AB^2 - AE^2 + 6CD$$ is:
Solution
Given that, $$AD\bot\ BC$$
$$BE=3$$
$$BC=5$$
Using Pythagoras' theorem,
$$AD^2+BD^2=AB^2$$ .....(1)
$$AD^2+DE^2=AE^2$$......(2)
$$AD^2+DC^2=AC^2$$......(3)
(1) - (2) gives
$$BD^2+DE^2=AB^2-AE^2$$
$$x^2-\left(3-x\right)^2=AB^2-AE^2$$
$$AB^2-AE^2=6x-9$$
$$AB^2-AE^2+6CD=6x-9+6\left(5-x\right)$$ ($$CD=\left(5-x\right)$$)
$$AB^2-AE^2+6CD=6x-9+30-6x$$
$$AB^2-AE^2+6CD=21$$
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