In an isosceles triangle ABC AB = BC, the median AD intersects the bisector CK at point O in this case.

| January 5, 2021 | education

In an isosceles triangle ABC AB = BC, the median AD intersects the bisector CK at point O in this case. AD perpendicular to CK Find the ratio: Sokbd: Sabc

In the AСD triangle, the segment CO is its height, since the ABP is perpendicular to the CК and the bisector, then the AСD triangle is isosceles, AС = AD. Since AM is the median of triangle ABC, then AC = CD = BD. Let AC = X cm, then BC = 2 * X cm.
The area of ​​the triangle ABC is equal to: Sавс = (AC * CB * sinC) / 2 = (2 * X² * sinC) / 2 = X² * sinC.
The area of ​​the triangle ADS is equal to: Sads = (AC * CD * sinC) / 2 = (X² * sinC) / 2.
Since CK is a bisector, then by its property: AK / AC = BK / BC.
AK / X = VK / 2 * X.
VK / AK = 2/1.
Since the triangles ВСK and AСK have a common height, the ratio of their areas is equal to the ratio of their bases.
Svsk / Sack = 2/1.
Sask = Svck / 2.
Savs = Svsk + Sack = Svsk + Svsk / 2 = 3 * Svsk / 2.
Svsk = 2 * Savs / 3.
Sokvd = Svsk – Ssod = (2 * Savs / 3) – (Savs / 4) = 5 * Savs / 12.
Sokvd / Savs = 5/12.
Answer: Squares are referred to as 5/12.



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