UPSC CDS 2 2021 Math Paper

Concept:Use Pythagoras theorem and the property that in an isosceles triangle, the altitude to the base is also the median.Explanation:In $\triangle ABC$, $AB = AC$. Draw $AE \perp BC$.Since $\triangle ABC$ is isosceles, $AE$ is also a median, so $BE = CE$.In right $\triangle ABE$: $AB^{2} = BE^{2} + AE^{2}$.In right $\triangle ADE$: $AD^{2} = DE^{2} + AE^{2}$.Subtracting the two equations: $AB^{2} - AD^{2} = BE^{2} - DE^{2}$.Since $BE = CE$, we have $AB^{2} - AD^{2} = CE^{2} - DE^{2}$.Factorise: $CE^{2} - DE^{2} = (CE - DE)(CE + DE)$.Observing the line segment lengths: $CE - DE = CD$ and $CE + DE = BD$ (depending on the order of points $D$ and $E$ on $BC$).Thus $AB^{2} - AD^{2} = CD \cdot BD$.Answer:Option A: $AB^{2} - AD^{2} = CD \cdot BD$.