Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 762.
MC cut AD at NTriangle MBc congruence to triangle MAN ... ( case ASA)so NA=BC=ADin right triangle NHD, HA is a median => AH=AN=AD
Extend DA and CM to meet at E.Triangles BMC and AEM are similarSo BC/AE = BM/MD = 1Follows AE = BC = ADTriangle HED is right angled at H and A is the midpoint of hypotenuse EDHence AH = AD (= AE)
Using vector approach to solve the problem. Let vector DA = a, vector DC = c.We have vector MC = 1/2 c - aLet vector DH = h, such that DH。MC = 0, => 1/2(h。c)= h。aNow |AH|。|AH|=(h-a)。(h-a) = |h|。|h| + |a|。|a| - 2(h。a) = |h|。|h| + |a|。|a| - (h。c)= |a|。|a| + |h|(|h|-|c|sinX), where X is the angle between DH and DC= |a|。|a|That means |AH| = |a| = |AD|Q.E.D.
Challenging problems in geometry (chaper 1, problem 16)...u.u
Let N be mid point of CD. MCNA is a parellllogram and AMHN is an isoceles trapezoid whose diagonals are equal MN = AH. But MN = AD since ADNM is a parallelogram and so AH = ADSumith PeirisMoratuwaSri Lanka