Given: ABCD – square;
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.
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