Maths Olympiad Prep

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Geometry Difficulty 6.2 National Olympiad Prove it JBMO

Problem:

The point PP is outside of the circle Ω\Omega. Two tangent lines, passing from the point PP, touch the circle Ω\Omega at the points AA and BB. The median AMA M, M(BP)M \in (B P), intersects the circle Ω\Omega at the point CC and the line PCP C intersects again the circle Ω\Omega at the point DD. Prove that the lines ADA D and BPB P are parallel.

Solution

Solution:

Since BAC=BAM=MBC\angle B A C = \angle B A M = \angle M B C, we have MABMBC\triangle M A B \cong \triangle M B C.

Figure 1

We obtain MAMB=MBMC=ABBC\frac{M A}{M B} = \frac{M B}{M C} = \frac{A B}{B C}. The equality MB=MP\quad M B = M P implies MAMP=MPMC\frac{M A}{M P} = \frac{M P}{M C} and PMCPMA\angle P M C \equiv \angle P M A gives the relation PMACMP\triangle P M A \cong \triangle C M P. It follows that BPDMPCMAPCAPCDAPDA\angle B P D \equiv \angle M P C \equiv \angle M A P \equiv \angle C A P \equiv \angle C D A \equiv \angle P D A. So, the lines ADA D and BPB P are parallel.

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