Question

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:

Accepted Answer

E. Let $BD = x \implies DE = 3 - x$ and $CD = 5 - x$. In right $	riangle ABD$, $AB^2 = AD^2 + x^2$. In right $	riangle ADE$, $AE^2 = AD^2 + (3-x)^2$. $AB^2 - AE^2 = x^2 - (3-x)^2 = 6x - 9$. $AB^2 - AE^2 + 6CD = 6x - 9 + 6(5-x) = 6x - 9 + 30 - 6x = 21$.

Triangle Perpendicular Side Relation

Explanation

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