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Let triangle $\triangle ABC$ have side lengths $AB = 7$, $BC = 8$, and $CA = 9$, and let M and D be midpoint of BC and the foot of the altitude from A to BC , respectively. Let E and F lie on AB and AC, respectively, such that $m\angle{AEM} = m\angle{AFM} = 90$ . Let P be the intersection of the angle bisectors of $\angle{AED}$ and $\angle{AFD}$ . Find MP.

We can easily calculate $BD=2$, $AD=3\sqrt5$, $MF=4/3 \sqrt 5$, $EM=12/7 \sqrt 5$.

Theoretically, massive coordinates calculation may work things out.

renmom
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  • See this. – user2661923 Nov 08 '20 at 00:23
  • I don't if this helps. But A,E,D,M,F are concyclic since all these points lie of the circle with diameter AE. MP is perpendicular to BC. – Isomorphism Nov 08 '20 at 03:12
  • I see A,E,D,M,F are concyclic in the circle with diameter AM. $\angle{PED} + \angle{PFD} = 45 $. How do you get $MP \perp {BC}$? Still don't know how to calculate PM. – renmom Nov 08 '20 at 06:12
  • Can you explain why ${MP} \perp {BC}$? If that's true, I can calculate $\tan \angle {DAM}$, which is the same as $\tan \angle {DFM}$, then I can get $\sin \angle {PFD}$. I can also get $\cos \angle {FMC}$ which is $\sin \angle {PMF}$. In $\triangle {PMF}$, I can get $PM$ using Law of Sine. – renmom Nov 08 '20 at 16:46
  • What's the source of the problem? – Dr. Mathva Nov 10 '20 at 11:45

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