Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1200.
http://s22.postimg.org/yor6ykls1/pro_1200.pngLet α= ∡ (CBM)= ∡ (PCB)= ∡ (BMA)And β= ∡ (NDC)= ∡ (DCQ)= ∡ (NAD)Triangle BCP congruent to tri. BAM …( hypoth- leg congruence)And Triangle DAN congruent to tri. DCQ …( hypoth- leg congruence)So AM=CP and AN= CQS(MAN)= ½. MA.NA.sin(MAN)= ½.MA.NA.sin(90- α- β)S(CPQ)= ½.CP.CQ.sin(PCQ)= ½.CP.CQ.sin(90+ α +β)Note that ∡ (MAN) supplement to ∡ (PCQ) so sin(MAN)= sin(PCQ) =>S(MAN)=S(CPQ)
Triangles BCP and ABM are congruent ASA and triangles ADN and CDQ are also congruent ASA.Hence <s PCQ and AMN can easily be shown to be supplementary.So altitudes QX and NY of Tr.s PCQ and AMN respectively are equal since Tr.s QCX and NMY are congruent ASASo using area = 1/2 base X heightS(CPQ) = S(AMN)Sumith PeirisMoratuwaSri Lanka
Is AM and AN perpendicular CP and CQ respectively.Therefore tri. ABM= tri. CBP or tri. AND=tri. CDQ (CP=AM ,CQ=AN, <PCB=<BAM, <DCQ=<NAD ) .But <PCQ +<MAN=180. Therefore (PCQ)/(MAN)=(CP.CQ)/(AM.AN)=1