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Let ABC be an acute-angled triangle in which ∠ABC is the largest angle. Let O be its circumcentre. The perpendicular bisectors of BC and AB meet AC at X and Y respectively. The internal bisectors of ∠AXB and ∠BY C meet AB and BC at D and E respectively. Prove that BO is perpendicular to AC if DE is parallel to AC.

 Let ABC be an acute-angled triangle in which ∠ABC is the largest angle. Let O be its circumcentre. The perpendicular bisectors of BC and AB meet AC at X and Y respectively. The internal bisectors of ∠AXB and ∠BY C meet AB and BC at D and E respectively. Prove that BO is perpendicular to AC if DE is parallel to AC.
 

Grade:11

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
Dear student

The solution is same just the variables are changed

Let ABC be an acute-angled triangle and suppose ∠ABC is the largest angle of the triangle. Let R be its circumcentre. Suppose the circumcircle of triangle ARB cuts AC again in X. Prove that RX is perpendicular to BC.
Solution: Extend RX to meet BC in E. We show that ∠XEC = 90◦ . Join RA, RB and BX.

Observe that ∠AXB = ∠ARB = 2∠C and ∠BXR = ∠BAR = 90◦ − ∠C.

Hence ∠EXC = 180◦ − 2∠C − (90◦ − ∠C) = 90◦ − ∠C.

This shows that ∠CEX = 90◦

Vikas TU
14149 Points
4 years ago
Let ABC be a triangle. Let X be on the segment BC such that AB = AX. Let AX meet the circumcircle Γ of triangle ABC again at D. Show that the circumcentre of 4BDX lies on Γ.
 Solution: Draw perpendicular from A to BC and extend it to meet Γ in F. We show that F is the circumcentre of 4BDX. Since AB = AX, we observe that F lies on the perpendicular bisector of BX. Join CF and CD. We observe that ∠ABX = ∠CDX and ∠AXB = ∠CXD. Hence 4ABX is similar to 4CDX. In particular 4CDX is isosceles. Moreover, ∠BCF = ∠BAF and ∠DCF = ∠DAF. Since AF is the perpendicular bisector of BX, it also bisects ∠BAX. It follows that CF bisects ∠DCX and hence F lies on the perpendicular bisector of DX. Together F is the circumcentre of 4BXD. 

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