Question

In ΔABC, AB = AC. Side BC is produced to D. Prove that `AD^2−AC^2`= BD.CD 

Answer 1

Draw AE⊥BC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:  

 

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle. 

So, BE = CE
And DE+CE=DE+BE=BD 

`AD^2=AE^2+DE^2` 

`⇒ AE^2=AD^2-DE^2  `       ...............(1) 

In ΔACE, 

`AC^2=AE^2+EC^2` 

⇒ `AE^2=AC^2-EC^2 `             ...............(2) 

Using (i) and (ii), 

⇒` AD^2-DE^2=AC^2-EC^2` 

⇒` AD^2-AC^2=DE^2-EC^2` 

                    `=(DE+CE) (DE-CE)` 

                     `(DE+BE) CD` 

                    `BD.CD`