Question

show that ,if two isocelies triangle have a common base?​

Answer 1

Given: ΔABC and ΔDBC are isosceles triangle with common base BC. AB = AC and BD = DC.

To prove: AD bisects BC at 90°.

Proof:

In ΔABD and ΔACD,

AB = AC           (Given)

BD = CD           (Given)

AD = AD           (Common)

∴ ΔABDΔACD     (SSS Congruence criterion)

⇒ ∠1 = ∠2        ...(1)     (CPCT)

In ΔABE and ΔACE,

AB = AC                                 (Given)

∠1 = ∠2                                  [Using (1)]

AE = AE                                  (Common)

∴ ΔABEΔACE                  (SAS congruency criterion)

⇒ BE = CD                ...(2)     (CPCT)

and ∠3 = ∠4                          (CPCT)

Now, ∠3 + ∠4 = 180°           (Linear pair )

∴ 2∠3 = 180°                         (∠3 = ∠4)       

⇒ ∠3 = 90°               ...(3)

Hence, AD bisects BC at 90°    [Using (2) and (3)]  

HOPE IT HELPS

Answer 2

Answer:

All isosceles triangles are not similar for a couple of reasons. The length of the two equal sides can stay the same but the measure of the angle between the two equal side will change, as will the base and the base angles.