Difference between revisions of "2002 JBMO Problems/Problem 1"
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− | Since <math>APBC</math> is a cyclic quadrilateral, <math>\angle CPA = \angle | + | Since <math>APBC</math> is a cyclic quadrilateral, <math>\angle CPA = \angle CPB</math>, and <math>\angle ABP = \angle ACP</math> --- (1) |
Now let <math>E</math> be the foot of the perpendicular from <math>C</math> to <math>AP</math>. Then we have, <math>CEPD</math> is a cyclic quadrilateral with <math>CP</math> as diameter of the circumcircle. | Now let <math>E</math> be the foot of the perpendicular from <math>C</math> to <math>AP</math>. Then we have, <math>CEPD</math> is a cyclic quadrilateral with <math>CP</math> as diameter of the circumcircle. | ||
− | It follows that <math> \triangle CPE</math> and <math> \triangle CDP</math> are congruent (since <math>\angle CPA = \angle | + | It follows that <math> \triangle CPE</math> and <math> \triangle CDP</math> are congruent (since <math>\angle CPA = \angle CPB</math>). |
So, we have <math>CE = CD</math> and <math>PE = PD</math> --- (2) | So, we have <math>CE = CD</math> and <math>PE = PD</math> --- (2) | ||
Latest revision as of 02:52, 6 June 2024
Problem
The triangle has
.
is a point on the circumcircle between
and
(and on the opposite side of the line
to
).
is the foot of the perpendicular from
to
. Show that
.
Solution
Since is a cyclic quadrilateral,
, and
--- (1)
Now let be the foot of the perpendicular from
to
. Then we have,
is a cyclic quadrilateral with
as diameter of the circumcircle.
It follows that and
are congruent (since
).
So, we have
and
--- (2)
Also, in we have
( from (1) above)
Thus
So from
,
is congruent to
. Hence we have
.
Now, (from (2) above)