In triangle ABC, Angle ACB=90°, seg CD is perpendicular to seg AB, seg DE is perpendicular to seg CD. Show that: CD² x AC = AD x AB x DE. Get link Facebook X Pinterest Email Other Apps September 30, 2018 Get link Facebook X Pinterest Email Other Apps Comments
In △BAC, Angle BAC= 90°, segment AD, seg BE and seg CF are medians. Prove: 2(AD²+BE²+CF²)=3BC² September 30, 2018 Read more
In the adjoining figure, AD is the bisector of the exterior angle A of triangle ABC. Seg AD intersects the side BC produced in D. Prove that: BD/CD=AB/AC. September 30, 2018 Read more
Comments
Post a Comment