Question 58
If D is the midpoint of side BC of a triangle ABC and AD is the perpendicular to AC then:
Solution
Using Apollonius theorem in $$\triangle$$ ABC, we can say that,
$$2(AD^2+BD^2) = AB^2+AC^2$$ ... (1)
In right-angle triangle ADC, $$DC^2 = AD^2 + AC^2$$ ... (2)
By equation (1) and (2), we can say that
$$2(DC^2 - AC^2 +BD^2) = AB^2+AC^2$$
$$3AC^2 = BC^2 - AB^2$$
Therefore, option A is the correct answer.
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