Consider triangles ABD and CBD: AB = BC (by condition), AD = BD (BD median), BD is common, triangles are equal, which means that their perimeters are also equal.
Let’s express the perimeter of each triangle:
P (ABD) = AB + AD + BD = 30 (cm),
P (CBD) = BC + BD + CD = 30 (cm).
Let’s add these perimeters:
P (ABD) + P (CBD) = AB + AD + BD + BC + BD + CD = AB + BC + (AD + CD) + 2BD.
Since AD + CD = AC, it turns out: P (ABD) + P (CBD) = (AB + BC + AC) + 2BD.
But (AB + BC + AC) is the perimeter of triangle ABC, it turns out:
P (ABD) + P (CBD) = P (ABC) + 2BD.
We substitute known data.
30 + 30 = 36 + 2BD.
2BD = 60 – 36;
2BD = 24;
BD = 24: 2 = 12 (cm).
Answer: BD = 12 cm.
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