Question

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). 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]

Answer 2

Answer:

BE =CF ... Proved.

Step-by-step explanation:

See the diagram given.

It is given that AC and AB are the equal sides in the isosceles triangle ΔABC.

Now, from equal sides AB and AC the altitudes CF and BE are drawn in the triangle ABC.

Therefore, area of ΔABC= 1/2(AC)(BE) =1/2 (AB)(CF) ....... (1)

{Since, area of a triangle = 1/2(Base)(Height)}

Since, AC=AB

Therefore, from equation (1), BE =CF ... Proved.