Depending on the situation you may discover useful result bearing such movement (which may be a rotation) of a line, an arc, a chord or a closed regular shape that quickly gives you the solution. This action of translation or movingof $\triangle EAD$ so that it is placed exactly on top of $\triangle ABC$ is a powerful technique of problem solving in Geometry. By this logic you can bypass showing line parallelism, a small step though. Second logicĪfter showing triangle congruency we can just superimpose $\triangle EAD$ on $\triangle ABC$ matching vertices so that ED merges with AC and automatically becomes perpendicular to CF. We have used problem breakdown technique in dividing the problem into two parts, first showing $AC \bot CF$ and then $ED || AC$ so that final result obtained is surely $ED \bot CF$. Key concepts used: Basic rhombus properties - triangle external angle concept - isosceles triangle property - congruent triangle tests - congruent triangle concepts - intersected parallel line concept. Result is, $\triangle AED$ is congruent to the $\triangle ABC$ so that $\angle ADE=\angle BCA = \angle x=\angle CAD$, as BC || AD. Next, as BC || AD, $\angle ABC = \angle EAD = \angle \alpha$ and the two pairs of sides that include this equal angle, AD=BC, and AE=AB.
$\angle \alpha = 180^0 - 2\angle x = 2\angle y$, as $\angle \alpha = \angle y + \angle y$, being the external angle in $\triangle CBF$, The following figure describes the problem, In a rhombus ABCD, AB is produced to F and BA is produced to E such that AB = AE = BF. Solutions to 10 difficult circle geometry and other geometry problems in SSC CGL Set 39 - answering time was 15 mins Problem 1. SSC CGL level Question Set 39, Geometry 7. Quick solutions use geometry problem solving techniques.įor greatest benefits take the test first at,
Circle geometry problems and solutions how to#
Learn how to solve quickly difficult circle geometry and other geometry problems in SSC CGL Set 39. Quick solutions to difficult circle geometry and other geometry problems
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