Question

In a triangle ABC, AB = AC and D is a point on side AC such that BC = AC $\times$ CD. Prove that BD = BC.

Answer 1

In order to prove the result, firstly draw $AE\perp BC$.

Perpendicular drawn from the vertex to the opposite base of an isoceles triangle bisects the base.
$\therefore BE=EC$
Applying Pythagoras Theorem in right $\Delta$AED,
$A{D}^2=A{E}^2+E{D}^2$ ...(1)
Again applying Pythagoras Theorem in right $\Delta$AEC,
$A{C}^2=A{E}^2+E{C}^2$ ...(2)
From (1) and (2), we get
$A{D}^2=(A{C}^2-E{C}^2)+E{D}^2$ $[A{C}^2=A{E}^2+E{C}^2\Rightarrow A{E}^2=A{C}^2-E{C}^2]$
$=A{C}^2+(E{D}^2-E{C}^2)$
$=A{C}^2+(ED+EC)(ED-EC)$
$=A{C}^2+(ED+BE)(ED-EC)$ $(\because EC=BE)$
$=A{C}^2=BD.CD$
Thus, $A{D}^2=A{C}^2+BD.CD$.