Question

$$\begin{gathered} \mathrm{In}△\mathrm{ABC},\mathrm{AB}=\mathrm{AC}\mathrm{and}\mathrm{D}\mathrm{is}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{BC}. \\ \mathrm{Prove}\mathrm{that}{\mathrm{AB}}^2-{\mathrm{AD}}^2=\mathrm{BD}.\mathrm{CD} \end{gathered}$$

Answer 1

In ΔABE and ΔACE,
$\angle$ABE =$\angle$ACE [AB = AC]
$\angle$BEA =$\angle$CEA [90$°$ each]
AE = AE [Common]
Therefore, by the AAS congruency test,
ΔABE$\cong$ΔACE

Thus,
BE = CE [Corresponding parts of congruent triangles]
Now, using the Pythagoras' Theorem in Δ ADE and Δ ABE, we get:
AD² = AE² + DE² …(1)
AB² = AE² + BE² …(2)
From (2) – (1), we get:
AB² – AD² = AE² + BE²– AE² – DE²
$\Rightarrow$ AB² – AD² = BE² – DE²
$\Rightarrow$ AB² – AD² = (BE + DE)(BE – DE)
$\Rightarrow$ AB² – AD² = (CE + DE)(BD) [BE = CE and BE – DE = BD]
$\Rightarrow$ AB² – AD² = (CD)(BD)
$\Rightarrow$ AB² – AD² = BD × CD

Hence proved.