AD, BE and CF are medians of a ∆ ABC. Prove that : 3(AB^2 + BC^2 + AC^2 ) = 4(AD^2 + BE^2 + CF^2)
Two right triangles ABC and DBC drawn on same hypotenuse BC & on same side of BC. Prove AP×PC=BP×DP.
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D and E are points on the sides AB and AC respectively of a △ABC such that DE||BC and divides △ABC
In Fig. the sides AB and AC of ∆ABC are produced to points E and D respectively. If bisectors BO
Sides AB and AC and median AD of a ΔABC are proportional to sides PQ,PR, PM of ΔPQR.Show ΔABC∼ΔPQR.
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Ex 6.3 QNo5 s and T are points on sides PR and QR of ΔPQR such that angle P = RTS. St. ΔRPQ~ΔRTS
O is a point in the interior of △ABC such that OA =12 cm OC = 9 cm∠AOB=∠BOC=∠COA and ∠ABC=6
In the given figure, if A B C ∼ D E F and the lengths of their sides (in cm ) are marked along th...
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ABC is a triangle. The bisector of the exterior angle at B and the bisector of C intersect at D. Pro
【高校数学】 数Ⅰ-93 三角形の面積① ・ 基本編
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