#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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.

Saurabh Koranglekar
one year 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
one year 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.