Question

ABC is and Isoceles triangle In which altitude’s BE and CF Are drawn two equal sidesAC And AB respectively ..Show that these altitudes are equal

Answer 1

ABC is an isosceles triangle     (given)  

BE is the altitude of AC             (given)

and CF is the altitude of AB      (given)

So,

$\angle{ABC}$ = $\angle{ACB}$   [As ABC is an isosceles triangle ]

$\angle{CFB}$ = $\angle{CFA}$ = $\angle{BEC}$ = $\angle{BEA}$      [as BE and CF are altitude of AC And AB respectively (so, each 90° ]

To be proof :

The altitudes BE and CF are equal.

Now,

In ΔAEB and ΔAFC

$\angle{BAE}$ = $\angle{CAF}$    (common)

$\angle{AEB}$ = $\angle{AFC}$     (each 90°)

AB = AC     [given (as ABC is an isosceles triangle) ]

Therefore

by AAS (Angle-Angle-side) congruence condition ΔAEB ≅ ΔAFC

Hence,

BE = CF     [by CPCT]