In the previous image and are bisectors and lies on the line segment . Prove that the distance from to line is equal to the sum of the distances from to lines and .
What happen if lies on the line but outside the triangle?
Now draw the circumcircle of triangle and suppose the ray intersects the circumcircle at ( is between and ). Prove that .
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Top NewestI proved The first 1... and based on that created a problem. Maybe if u likeJorge,,,, you may share this problem. So here's the problem. In an acute angled triangle ABC we have BC=4,AC=5,AB=6. BM and CN are the internal bisectors of B and C. P is any point on the segment MN. If [△PMC]=1 and [△PNB]=4 , find [△BPC] . ( it wud be nice if u mention a little credit to me.. :P )
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Reply if you find this problem interesting! :)
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I think there are too many conditions in your problem. Triangle ABC is fixed and also the points M and N, thus if you choose P in line segment MN such that [BPN]=2 then [CPM] is determined and is not equal to 4. Another idea is to calculate [MNBC] and then find [BPC] (I think the answer will be different of yours).
Try this problem (using your notation):
If [BPN]=2, what is the correct value of [CMP]?
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Then cant we fix P and say that suppose there exists a point P on MN such that ........ ?
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However, solution to #1. Let PX⊥BC, PY⊥AC and PZ⊥AB. We drop MM1⊥BC, MZ1⊥AB and NN1⊥BC and NY1⊥AC. Now we have △BM1M≅△BMZ1 (right angle, angle and side equal) and hence MZ1=MM1. Similarly we have, NN1=NY1.
We use similarity in triangles MPY and MNY1.
NY1PY=NMMP and we get PY=NMPM×NY1...(i) Similarly we have PZ=NMNP×MZ1...(ii)
Now, NN1,PX,MM1 are 3 parallel lines . So we have NMPM=M1N1M1X.... (iii) We join N,M1 and let that intersect PX at L.
Using △NLP∼NM1M , we have MM1PL=NMNP and thus PL=NMNP×MM1=NMNP×MZ1. (as MM1=MZ1 ... (iv)
And, similarly using △M1XL∼△M1N1N , we have
LX=N1M1XM1×NN1=NMPM×NY1...(v)
We observe that (iv)+(v) is indeed (i)+(ii) and Hence PY+PZ=PL+LX=PX
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Is it correct Jorge??
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It seems good, but you misplaced the points, for example you said MM1 perpendicular to BC.
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