Question

Abc and Dbc are two isosceles triangle on the same base BC show that angle ABD= angle ACD​

Answer 1

Answer:

Well..the question can have 2 or more solutions. One of them would be..

Step-by-step explanation:

So, first in the figure, we have triangles ABC and DBC as isosceles (given)

drawing a straight line AD from angle A to D.

Now, we've two more triangles I.e. triangle ABD and ACD.

In triangle ABD and ACD, AB = AC (given)

BD = DC (given)

AD = AD (common)

Therefore, triangle ABD is congruent to triangle ACD.

So, angle ABD = angle ACD (CPCT)

OR..

Because angle ABC = angle ACB (isosceles triangle)[GIVEN]

and angle DBC = angle DCB (isosceles triangle) [GIVEN]

therefore, angles ABC + DBC = ACB + DCB

so, angle ABD = ACD (Hence, proved)

Answer 2

GIVEN:-

ABC and DBC are isosceles ∆

BC = BC

TO PROVE:-

$\large\sf{\angle{ABD}=\angle{ACD}}$

PROOF:-

$\large\sf{AB=AC}$

$\large\sf{\angle{ABC}=\angle{ACB}}$.......(i)

$\large\sf{BD=CD}$

$\large\sf{\angle{DBC}=\angle{DCB}}$........(ii)

$\huge\sf\red{NOW,}$

$\large\sf{\angle{ABC}+\angle{DBC}=\angle{ACB}+\angle{DCB}}$

$\large\sf{\angle{ABD}=\angle{ACD}}$

$\huge\sf\orange{Hence\:Proved}$