Question 56

Ina $$\triangle$$ABC, AD is perpendicular to BC from A. if $$\angle BAC = 90^\circ, then AB^2 : AC^2$$ is equal to:

Solution

AD is prependicular to BC and Angle BAC = 90

Then $$AD^2=BD\times\ CD$$

And,

$$AB^2=BD^2+AD^2$$ = $$BD^2+BD\times\ CD$$ = $$BD\left(BD+CD\right)$$----------I

$$AC^2=CD^2+AD^2$$ = $$CD^2+BD\times\ CD$$ = $$CD\left(CD+BD\right)$$--------II

Dividing eqn I by II,

$$\frac{AB^2}{AC^2}=\frac{BD}{CD}$$

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