Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 2"
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==Solution== | ==Solution== | ||
− | Since <math>\triangle BDC</math> is isosceles, <math>BD=DC</math>. Since <math>\angle ABD = \dfrac{\pi}{2}</math>, <math>\dfrac{\pi}{2} = \angle ABD = \angle ABC + \angle DBC = \angle ACB + \angle DCB = \angle ACD</math>, which means that <math>\angle DCF = \dfrac{\pi}{2}</math>, too. Thus <math>\angle EBD = \angle DCF</math>, so by ASA, <math>\triangle EBD \cong \triangle FCD</math>. This means that <math>ED = FD</math>. Since <math>AB = AC</math>, <math>BD = DC</math>, and <math>\angle ABD = \angle ACD</math>, by SAS, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle BDA = \angle ADC</math>. Since <math>\angle BDE = \angle MDP</math>, <math>\ | + | Since <math>\triangle BDC</math> is isosceles, <math>BD=DC</math>. Since <math>\angle ABD = \dfrac{\pi}{2}</math>, <math>\dfrac{\pi}{2} = \angle ABD = \angle ABC + \angle DBC = \angle ACB + \angle DCB = \angle ACD</math>, which means that <math>\angle DCF = \dfrac{\pi}{2}</math>, too. Thus <math>\angle EBD = \angle DCF</math>, so by ASA, <math>\triangle EBD \cong \triangle FCD</math>. This means that <math>ED = FD</math>. Since <math>AB = AC</math>, <math>BD = DC</math>, and <math>\angle ABD = \angle ACD</math>, by SAS, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle BDA = \angle ADC</math>. Since <math>\angle BDE = \angle MDP</math>, <math>\angle EDA = \angle BDA - \angle BDE = \angle ADC - \angle MDP = \angle CDP</math>. Thus <math>\angle EDP = \angle EDA + \angle MDP = \angle PDC + \angle CDF = \angle PDF</math>. <math>DP</math> is the angle bisector of <math>\angle EDF</math>, and <math>ED = DF</math>. This means that <math>EF \perp DP</math>. <math>AM = AM</math>, <math>BM = MC</math>, and <math>AB = AC</math>, so by SSS, <math>\triangle ABM \cong \triangle ACM</math>. Thus <math>\angle AMC = \dfrac{\pi}{2}</math> and <math>\angle DMP = \dfrac{\pi}{2}</math>. <math>\angle MDP = \angle BDE</math>, so by AA, <math>\triangle BDE \sim \triangle MDP</math>. Thus <math>\dfrac{ED}{BD} = \dfrac{DP}{MD}</math>. Also, <math>\angle EDF = \angle EDC + \angle CDF = \angle EDC + \angle EDB = \angle BDC</math>. <math>BD = DC</math> and <math>ED = DF</math>, so <math>\dfrac{BD}{ED} = \dfrac{DC}{DF}</math>. By SAS similarity, <math>\triangle BDC \sim \triangle EDF</math>. <math>MD</math> is a median and an angle bisector of <math>\triangle BDC</math>. Now assume that P' is the point such that DP' is a median of <math>\triangle EDF</math> (it is on <math>EF</math>). It is on DP, the angle bisector, and since <math>\triangle BDC \sim \triangle EDF</math>, <math>\dfrac{ED}{BD} = \dfrac{DP'}{MD}</math>, but we also showed that <math>\dfrac{ED}{BD} = \dfrac{DP}{MD}</math>. Thus <math>DP' = DP</math>. Since P and P' are on the same ray (<math>DP</math>), P = P' and P is the midpoint of <math>EF</math>. |
==See also== | ==See also== |
Latest revision as of 14:05, 15 December 2022
Problem
Let and
be isosceles triangles with the base
. We know that
. Let
be the midpoint of
. The points
are chosen such that
,
,
, and
. Prove that
is the midpoint of
and
.
Solution
Since is isosceles,
. Since
,
, which means that
, too. Thus
, so by ASA,
. This means that
. Since
,
, and
, by SAS,
, so
. Since
,
. Thus
.
is the angle bisector of
, and
. This means that
.
,
, and
, so by SSS,
. Thus
and
.
, so by AA,
. Thus
. Also,
.
and
, so
. By SAS similarity,
.
is a median and an angle bisector of
. Now assume that P' is the point such that DP' is a median of
(it is on
). It is on DP, the angle bisector, and since
,
, but we also showed that
. Thus
. Since P and P' are on the same ray (
), P = P' and P is the midpoint of
.