Question

124
A
7. 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

Answer:

ABC is an isosceles triangle (given)

BE is the altitude of AC (given)

and CF is the altitude of AB (given)

So,

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

\angle{CFB}∠CFB = \angle{CFA}∠CFA = \angle{BEC}∠BEC = \angle{BEA}∠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}∠BAE = \angle{CAF}∠CAF (common)

\angle{AEB}∠AEB = \angle{AFC}∠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]

HENCE PROVED