SOLUTION: Suppose that M lies on the side BC of triangle ABC, and angle BAM = angle CAM = a. Let the line through C parallel to MA meet BA produced at P.
Prove that triangle APC is isosce
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-> SOLUTION: Suppose that M lies on the side BC of triangle ABC, and angle BAM = angle CAM = a. Let the line through C parallel to MA meet BA produced at P.
Prove that triangle APC is isosce
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Question 1113548: Suppose that M lies on the side BC of triangle ABC, and angle BAM = angle CAM = a. Let the line through C parallel to MA meet BA produced at P.
Prove that triangle APC is isosceles and hence AP = AC Found 2 solutions by mananth, greenestamps:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! .
MA is parallel to PC
and
AC is the transversal
Therefore CAM = angle ACP ( alternate angles)
MA is parallel to PC
and
BP is the transversal
BAM =APC ( corresponding angles)
But BAM = CAM =a
therefore APC = ACP
Therefore CAP is an isosceles triangle
hence AP = AC
A different approach than used by the other tutor....
(1) angle MAC is congruent to angle ACP -- alternate interior angles
(2) angle BAM is congruent to angle CAM -- given
(3) angle PAC plus angle BAC = 180 degrees -- straight angle
(4) angle PAC plus angle ACP plus angle CPA = 180 degrees -- sum of angles of a triangle
(5) angle APC is congruent to angle ACP -- (3) and (4)
(6) triangle APC is isosceles -- two congruent angles
(7) AC = AP -- sides opposite congruent angles in an isosceles triangle
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