ABCD – square a) Find the perimeter of the triangle b) Find the area of the triangle c) Prove that triangle ABD = triangle BCD

Given: ABCD – square;
To find:

a) PABD -? cm;

b) SABD -? cm;

c) Prove that tr-k ABD = tr-ku BСD.

a) PABD = AB + BD + AD. The sides of the ABD track are also the sides of the ABCD square. Because all sides of the square are equal, so tr-k ABD is isosceles, and AB = AD, i.e. PABD = 2 * AB + BD.

b) SABD = SABCD: 2; SABCD = (AB) ^ 2.

So SABD = (AB) ^ 2: 2.

c) R / m tr-k ABD and tr-k BСD.

In these treugs. BD is the common side. Because ABCD is a square, then AB = AD = DC = BC.

So tr-k ABD = tr-k BСD.