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Archive for the ‘Plane_Geometry’ Category

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…

Triangle problem – finding height with given area and angles.

If the area of triangle ABC is ${\cal S}$ and angles $\angle A= \alpha$ and $\angle B= \beta$ then find a altitude of triangle which drawn from C to AB.

Categories: Triangles Tags: , , ,

Triangle problem – find an angle

Triangle ABC is isosceles, AC=BC. $\angle ADE=10^\circ , \angle CBD=20^\circ .$ Find measure of angle $\angle AED$ .

SOLUTION:

Categories: Triangles Tags: , , ,

Triangle – Area problem

In triangle ABC, cevians AD, BE and CF intersect at point P. The areas of triangles PAF, PFB, PBD and PCE are 40, 30, 35 and 84, respectively. Find the area of triangle ABC.

Categories: Triangles

Triangle Problem

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.

Categories: Triangles Tags: , ,

Triangle problem

October 4, 2011 1 comment

Let triangle ABC be acute and let H be its orthocenter. The altitudes $AA_1 \: BB_1 \: CC_1$ . Prove that $\frac{AH}{AA_1} + \frac{BH}{BB_1} + \frac{CH}{CC_1} = 2$
It is easy question. Use areas to solve it.

Categories: Triangles Tags: , , ,

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 $p \cdot q = \frac{ad+bc}{ab+cd}$ .
It is easy question, just use $R = \frac{abc}{4S}$ a, b, c are sides of triangle and S is area of triangle.