Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, December 8, 2016

### Geometry Problem 1293 Triangle, Altitudes, Two Squares, Center, Concurrent Lines, Midpoint

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Connect OA and OB

ReplyDeleteObserve that qua. OBHA is cyclic

So ∠BAH=∠BOH

In triangle ABH , ∠BAH complement to ∠B => ∠BOH complement to ∠B

So ∠BOH supplement to ∠OBF => OH //BF

Since O is the midpoint of EB and OH//BF=> OH cut EF at midpoint M

Similarly QJ also cut EF at midpoint M

Quad. AOBH is cyclic,Angle BOH= Angle BAH =90-B,Angle OBQ=90+B,Hence OH is parallel to BF. Since O is midpoint of BE, line OH will pass pass through midpoint of EF.

ReplyDeleteSimilarly QJ is parallel to BE and passes through midpoint of EF. Hence QJ, OH and EF are concurrent at mid point of EF.

m(BHA) = m(BOA) = 90 => OABH is cyclic quadrilateral

ReplyDeletesince m(BAO) = 45 => m(BHO) = 45 => m(OHA) = 45 --------(1)

Extend AH to meet BF at P and consider the triangle BAP

we have

m(ABP) = B+45

m(BAP) = 90-B

=> m(BPA) = 45 ----(2)

From (1) and (2), we can say OH||BF

Since O is mid-point of BE, from mid-point theorem, M is the midpoint of EF

similarly QCJB can be found cyclic and thus JQ intersects EF at M