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Archive for the ‘Plane_Geometry’ Category

Quadrilateral Problem

January 7, 2013 Leave a comment

In quadrilateral ABCD with diagonals BD and CA, \angle ABD = 40^{ \circ } , \; \angle CBD = 70^{ \circ } , \; \angle ABD = 40^{ \circ } , \; \angle CDB = 50^{ \circ } , \; \angle ADB = 80^{ \circ } . \; Find \; \angle CAD . Read more…

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Triangle problem – finding height with given area and angles.

January 4, 2013 Leave a comment
Categories: Triangles Tags: , , ,

Triangle problem – find an angle

December 16, 2011 Leave a comment
Categories: Triangles Tags: , , ,

Triangle – Area problem

October 31, 2011 Leave a comment

Triangle Problem

October 8, 2011 2 comments

Let triangle ABC be isosceles with AB = AC. The altitude from A is AE and cevian BF intersects AE at D. If AF : AC = 1 : 3 Then find AD : DE and BD : BF.

 

Just draw cevian CG that intersects AE at D, too . (G is on line AB) Connect F & G. Line FG is parallel to BC. Use similarity of triangles to solve the questions.

ANSWERS ARE Read more…

Categories: Triangles Tags: , ,

Triangle problem

October 4, 2011 1 comment
Categories: Triangles Tags: , , ,

Quadrilateral & Circle problem

October 1, 2011 3 comments
medival ideal portrait of Ptolemy

Image via Wikipedia

Quadrilateral ABCD is inscribed in a circle. Let AB = a, BC = b, CD = c, DA = d, AC = p and BD = q. Prove Ptolemy’s second theorem that p \cdot q = \frac{ad+bc}{ab+cd} .

This question is also taken from Bertley Math Circle .

It is easy question, just use R = \frac{abc}{4S} a, b, c are sides of triangle and S is area of triangle.

If you couldn’t prove send me mail, and i will give few tips.

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