Question

Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal

Answer 1

Answer:

the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal

Step-by-step explanation:

Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal

Let say ABC  is  IsoscellesTraingle

where BC is the Base

& AB = AC  equal sides

Let say BD ⊥ AC    & CE⊥AB

Area of Triangle = (1/2)AC  * BD     & (1/2)AB * CE

=> (1/2)AC  * BD    = (1/2)AB * CE

=> AC * BD = AB * CE

AC = AB

=> BD = CE

Hence the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal

Answer 2

Proof:

In ∆ABC,  AB = AC [data]

∠ABC = ∠ACB  

In ∆EBC and ∆DCB

∠EBC =∠DCB( Base angles )  

∠BEC = ∠CDB [= 90° ]  

BC = BC (Common side)  

∆EBC = ∆DCB [ASA postulate]  

BD = CE [Corresponding sides]