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Given ∆ ABC right angled at B and AD, CE are two medians drawn from A & C respectively on sides BC & AB. If AC = 5cm & AD = 3√5/2
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In ∆ABC, AB=10cm BC=15cm. A line is drawn from the vertex B to AC such that AD:DC=2:3. Angle(BDC)=130° Angle(BCA)=30°. Then, Angle(ABC) is equal to (A) 30° (B) 40° (C) 45° (D) 110°
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Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect
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